I tnuat gtat ttnann m iisi vat L t i^, tap I - Nguyen

Quang

Lac

A? + A^ + A^ => A2 =

Vai may FX570-ES ta bam n h u sau :
[ M O D E J § t r e n m a n h i n h x u a t h i # n c h u : CMPLX

,y

Mat khac:

- A ^ = V42

-2^ = 2V3 (cm)

A^ = A^ + A^ + 2A^Aj cos(92 - 9])

|SHIFTJ|MOD£| § chuyen doi don vi do goc la R (Rad):
fi!

|4>/3|^HIFTJ 0. § g § ISHIFTJ 0
[SHIFT] 0

§ 0

'^^'^ ^^^^

H

•

'^^t

•

o

j h li jisv'v' :jr.

2^ =4^+(2V3)%2.4.2N/3COS

4 Z 7t/2

cos

Nghia la bien do bang A = 4; pha ban dau cf) = n/2
=> Chon A
Cau 146. Mot vat thuc hi^n dong thoi hai dao dpng dieu hoa cung phuong c6
phuong trinh dao dong la X j = 4 s i n ( 7 t t + a) (cm),

= 4\/3cos7ct (cm). Bien do

371

2

D. a = T:


C.a = 0

Huong dan gidi:
Ta c6: x, = 4 7 2 cos

Ttt

+ a — (cm) va X j =4\/2cos(27tt) (cm).
2

Bien dp dao dong tong hop xac djnh:
+2AjA2COS a —
2

De bien do dao dong dat gia tri nho nha't thi

cot-^

3j

cm; X2 = A2cos(o)t + (p2)cni

iML.'''i^..\

B. Tan so ciia goc dao dpnh tong hpp co = 27t rad/s

,.

;


C. Pha ban dau cua dao dong tong hpp 9 = —
4
D. Phuong trinh cua dao dpng tong hpp x = 8cos

4

cm

Huong dan gidi:
Taco: x, = 4 V 2 s i n 2 7 i t ( c m ) va X j = 4\/2cos27tt (cm)

A

Bieu dien cac dao dong tren gian do vec to,
ta c6: OA = A i = 4cm; OB = A2;. OC = A = 2cm

dap an B d i i n g .

2
2
2
A 2 - A ^ + A ^ + 2 A J A 2 C O S ( 9 2 - 9 I ) = (4V2) + ( 4 V 2 ) + 2 ( 4 7 2 ) . c o s ^
=> A = 8cm => Dap an A diing.
Pha b a n d a u c u a d a o d p n g tong hpp d u p e x a c d i n h :
A , sin(p, + A , sin(p,

— = -1:

.(p =


—

dap an C diing.

jf'.

A^ C O S ( p j + A2 COS(p2

C. sVScm; 0

Huong dan gidi:

BC2 = O C + OB2

A . Bien dp cua dao dpng tong hpp la A = 8cm

tgcp = — •

Cap gia tri nao cua A2 va cp sau day la dung?

T u gian do vec to suy ra:

0ii_148. M o t vat thuc hien dong thoi 2 dao dong c6 phuung trinh la

Bien d p c i i a d a o d p n g tong hpp x a c d j n h :

Phuong trinh dao dong tong hop x = 2cos((ot + (p)cm. Trong do cpj -(p = --

B. 2V3cm; 4


'.....VP.

Vi h a i d a o d p n g t h a n h phan deu co tan so goc co = 27irad/s nen d a o d p n g

Cau 147. Mot chat diem tham gia dong thoi hai dao dpng dieu hoa tren cung

A. sVScm; 2

0

2

tong hpp cOng co tan so goc (a = 2n rad/s

cos a — = - l = > a - — = 7t=:i>a = — ^ Chpn A
2
2
2

xi = 4cos

3;

Vay A2 = 2\/3(cm); (p = 0 => Chpn D

n

A^ = A j + A j + 2 A j A j cos((p2 -(pj) = A j + A j


mot true Ox c6 phuong trinh:

•cp =

6J

I

X j = 4^2 sin27it(cm); X2 = 4>/2 cos27it (cm). Ket luan nao sau day la sai? , >

dao dong tong hop dat gia tri nho nha't khi
A. a =

571

71

n

^T^^->

D. 2V3cm;0

V5 /

Phuong trinh dao dpng dieu hoa tong hpp co bieu thuc:
X = 8cos 27rt-i^ c m
4

d a p a n D s a i -> C h p n D


V o i may FX570-ES ta bam n h u sau :

7-.

P O D E ] ^ tren man hinh xuat hien chu: CMPLX
| H I F T ] [ M O D E J § chuyen doi don vi do goc la R (Rad):

f72

SHIFT

(-) -7t/2

SHIFT

(-) 0
161


-

SHIFT] y

g y

M a n h i n h hien t h i ket qua: 8 Z - 7 t / 4

A=(V3B)^-4(52-25)>0


NghTa la bien d o bang A = 8; pha ban dau 4> = - H / 4
A d i i n g ; B d i i n g ; C d i i n g ; D sai. ^ C h o n D .

Bmax = lO(cm) thay vao (1) ta dupe A = S^/Scm => C h p n A

C a u 149. M g t vat t h a m gia d o n g t h a i hai dao d o n g c i i n g p h u o n g , cung tan so
C O bien d o Ian lug-t la A i = 3cm va A 2 = 4cm. Bien do dao d g n g t o n g h o p khorig
the nhan gia t r i nao sau day:
A . 5,7cm

B. 1,0cm

C. 7,5cm

Huong dan

M p t con lac 16 xo t h a m gia d o n g t h o i 2 dao d p n g d i e u hoa c i i n g
phuong, c i i n g tan so' co = 5\f2 rad/s, c6 d p I f ch pha bang ^ . Bien d p ciia 2 dao
dpng thanh p h a n la A i = 4 c m va A 2 . Biet d p I o n van toe ciia v a t tai t h o i d i e m

D . 5,0cm

(Jpng nang bang the nang la 40( cm). Bien dp thanh p h a n A 2 bang

gidi:

Theo p h u o n g phap gian d o vec to Fre-nen t h i bien d o dao d p n g tong h o p

A.


A2

= 4 cm

B.

A2 = 4V3

l u o n phai thoa m a n : | A j - A j | < A < A j + A j . Theo bai ra, ta c 6 : 1 c m < A < 7cm.

D . ±407i cm/s

(0

COS(A(P)

= 75 => A = 5 V 3 c m

a'

V = ±4071 ( c m / s)

gidi:

• A = 8cm

X 2 = Bcos c o t - ^ cm (t d o bang giay). Bie't p h u o n g t r i n h dao d p n g tong
2

hop


C a u 153. Cho hai dao d p n g d i e u hoa c i m g p h u o n g : X j = 2 c o s ( 4 t + (pj)cm;
= 2 c o s ( 4 t + (p2)cm v o i 0 < (P2 -(Pj ^ n. Biet p h u o n g t r i n h dao d p n g tong
x = 2cos

hop

la X = 5cos(cot + (p)(cm). Bien dp dao dpng B c6 gia tri cue dai k h i A bSng:
B. 5 cm

,

C. Syflcm
Hwang dan

D . 2,5V2cm

Tir cong thiic tong h o p dao dpng ta c6:
571

6

c m . Gia t r j ciia cpt la:

6
'ff':

B.

66


'
Huang dan

2
gidi:

Gian d o vecto n h u h i n h ve. Ox la tri^c goc:

gidi:

5^ = A ^ + B^ - 2 A B C O S — hay A ^ - VSBA + B^ - 2 5 = 0
6
De p h u o n g t r i n h (1) c6 n g h i ^ m A v o i B la t h a m so t h i :

Al=A^-Al=8^-A^=>A2=4Scm

=> C h p n B

Chpn D

CO

^ n
C a u 151. Cho h a i dao d p n g dieu hoa c i i n g p h u o n g : X j = A c o s a->t + — cm vn
3

A . 573cm

= 6 cm


A ^ = A J + A ^ + 2A1A2 C O S ( A 9 = A ^ + A ^ )

^

CO

A2

gidi:

A p d y n g cong thuc:
v2

D.

M a t khac ta co:

A^ = A^ + A^ + 2A1A2

2

2v^

A^=2--

Ta c6: co = 27:f = 871 (rad/s)

,2


= \/3 c m

E = 2W^ = i k A 2 = > 2 - m v 2 = i k A 2
d
2
2
2

Van to'c ciia vat k h i no c6 gia toe 32>/2 (cm/s^) la
Huong dan

A2

K h i d p n g nang bang the nang, ta co:

c i m g tan so' i = 4 H z , c i i n g bien d o 5 cm va c6 d o lech pha A(p = —. Cho 7t^ = 10.

C. ± 3 0 ^ 2 cm/s

C.

Ta C O co nang ciia vat: E = W j + W, = ^ k A ^

C a u 150. M o t vat t h y c hien d o n g t h o i hai dao d o n g d i e u hoa c i i n g phuong,

B. ± 2 0 ; : cm/s

cm

Huang dan


Vay A k h o n g the bang 7,5cm => Chpn C

A . ±307c cm/s

<=>B<10(cm)

OA = O A i + O A 2

(1)

Theo bai: O A i = O A 2 = 2 (cm)
=> T a m giac O A A 2 can tai O => O A A 2 = O A 2 A
O A la p h a n giac A j O A 2 ( V i O A 2 A A 1 la h i n h thoi).


Q 2- Thai gian chuyen dpng thling cua v|t m t u liic ban dau den v i tri 16 xo

AOA2 = AOAj = OAA2

l^'hong bien dang la:

=> A O A , = O A A 2 = O A 2 A
A A O A 2 deu ^

',

A O A 2 = - =i> AOA,
2
3


A.

= -

' 3

Theo bai: AOx = — => A^Ox = — =>(p, = - —
6
^
6
^
6

kx = fjmg => X = |Limg/k = 2 (cm).

giay. Biet dp Ian van toe ciia vat tai thoi diem dpng nang bang the nang ij

,^ B. 2V3cm

C. 373cm

D. 4V3cm

Thai gian chuyen' dpng thSng ciia vat m t u liic ban dau den v i tri 16 xo
'

^ ;

rSu 157. * Mot con lac 16 xo gom mot vat nho kho'i lupng 0,02kg va c6 dp cung


Ion nhat vat nho dat dupe trong qua trinh dao dpng la
A. 1572cm/s

• A = 8cm

B. 25V2em/s

C. 40^^ cm/s

D. 50v^em/s

Huang dan gidi:

Do hai dao dpng thanh phan vuong pha nen

Van toe ciia vat dat gia trj Ion nhat trong qua trinh dao dpng tat dan ehinl"

A 2 = ^A^ - A j = 78^-4^ = 4V3cm => Chpn D

v$n toe Ian nhat ma vat dat dupe trong i chu ki dau tien. Gpi x la v i tri l i do

Cau 155. Mot con lac 16 xo dao dpng tat dan. Cu sau moi chu ki bien dp ciia no
giam 2,5%. Phan nang lupng cua con lac bj mat d i sau moi dao dpng toan
phan la
' '^

•-

bj nen 10 cm roi buong nh^ de con lac dao dpng tat dan. Lay g = lOm/s^. Toe dp


2imv2=ikA2

^

"^'•>

H? so ma sat trupt giiJa gia da va vat nho la 0,1. Ban dau g i u vat 6 v i t r i 16 xo

Khi dpng nang bang the nang ta c6: 2Ed = E

A. 35%

T = 2 7 t J ^ - 0,2n (s)

1 N/m. Vat nho dupe dat tren gia do c6 dinh nam ngang dpe theo trvic 16 xo.

Huang dan gidi:

k

Chu ki dao dpng:

khong bien dang la: t = j + - ^ = -^(s) => Chpn C

40cm/s. Hay xac djnh bien do dao dpng thanh phan Aj ?
,

D. ^- s.
30


C.^s.
15

Vi tri can bang ciia con lac 16 xo each v i tri 16 xo khong bien dang x:

Chpn B

=4cos^5>/2t-njcm va X 2 = A 2 C O S 5\ll\ — c m , trong do t tinh bang

A. VScm

•

Huong dan gidi:

7t

Cau 154. Mgt vat tham gia dong thai hai dao dong dieu hoa cung phuang v6,
Xj

B.^s.
20

25S

B. 45%

C. 55%


D.95%

Huang dan gidi:
1

2 2

Nang lupng trong chu ki dau la: W = - mco A
Nang lupng trong chu ki tiep theo la: W ^mco^A'^
A/
= (0,975)^ * 0,95 = 95% =^ Chpn D
Suy ra:
w
A
Cau 156. * Mot con IMc 16 xo c6 dp ciing k = 10 (N/m), khoi lupng vat nan?
m = 100 (g), dao dpng tren mat phang ngang, dupe tha nhe tu vi tri 16 xo d^^
6cm so voi v i t r i can bang. H? so ma sat trupt giua con lac va mat ban bang H *

ma vat c6 toe dp Ion nhat.
Ap dung dinh luat bao toan nang lupng, nang lupng ban dau ciia h§ gom the
nSng cue dai ban dau: ~ k A p , nang lupng ciia h§ tai thoi diem vat c6 van toe c i . ;
?i gom dpng nang ^ m v ^ , the nang ^^x^ va dp Ion eong ciia lye ma sat:
= ^mg Al

= |img(Al + x)

1 kx^ +)img(Al + x)
-kAl^ =-mv^ + —
. 2 ^ J i A l 2 - A x 2 - 2 ^ g ( A l + x)
m

V
/
m
Dat y = v 2 = — A l ^ - — x 2 - 2 ^ g ( A l + x ) . X e t d a o h a m y' = - — x - 2 ^ i g
m m
m
y' = 0 » X = - 1 ^ ^ = -0,02m = -2em

'

.

,,


Ki thuatgilii

iiluiiih

BI'rN

Vat Li 12, t&p 1 - Nguyen Quang Lac

Cfy TNHH

K h a o sat svt bien thien ciia y ta tha'y v o l x = - 2 c m t h i y dat gia t r j cue

{Chi vat d u n g h i n t h i co nang cua vat chuyen het thanh cong ciia l y c m a sat.

Suy ra v dat gia t r j eye dai tai v i t r i x = - 2 c m


Do do: A ^ ^ = Hmg.s = W => 0,02.0,1.10.5 = 0,5 => S = 2 5 ( m )
=>ChpnB

v ^ 3 ^ = J — A l ^ - — x 2 - 2 n g ( A l + x) = 0 , 4 V 2 m / s = 40N/2cm / s

' '

£^vi 161. M p t con lac lo xo n a m ngang d a n g dao d p n g tat dan. N g u o i ta d o

=> C h g n C

jju<7C d p g i a m t u o n g d o ! ciia bien d p trong 3 ehu k y d a u tien la 8%. D p g i a m

Cau 158. M p t con lac 16 xo n a m ngang g o m v ^ t n h o kho'i l u g n g 200 (g), 16 xo
d p Cling 10 ( N / m ) , h$ so m a sat t r u p t giua v$t va m a t phSng ngang la 0,1. Ba^
d a u vat d u p e g i i i 6 v i t r i 16 xo gian 10 cm, r o i tha nh? de con lac dao dpng tjj

tuong d o i ciia the'nang dan hoi t u o n g u n g
A. 5%

B. 10,3%

= 0,92, suy ra ^

ciia vat bat d a u g i a m t h i d p g i a m the'nang ciia con lac la:
B. 24mJ.

C 36 mJ.


C. 15,4%

D . 23,1%

Huang dan gidi:

dan, lay g = 10 (m/s^). T r o n g khoang t h o i gian ke t u liic tha cho den k h i toe ^6
A . 12mJ.

Kangvi^

Ml

= 100% - 8% = 92%

^0

D.48mJ.

Huang dan gidi:

Ta lai c6:

Vgt dat v a n toe cue dai k h i Fdh = Fms

Hay ^

Wn

=> kx = ^ m g => X = ^^^^ = 2 (cm)

D o d o d o g i a m the nang la: AW, = - (A^ ) = 0,048 J = 48 m j .

AW,

W(, - W.1

w„

Wn

w.
VV-i

n-n

^ = 1 -W,^ = 1-—5-

1

= 1 - 0 , 9 2 ^ = 0,154

= 15,4% ^ C h o n C

Cau 162. * M p t con lac 16 xo dat tren mat phang n a m ngang, 16 xo c6 d p cung
10(N/m), vat nang co k h o i l u p n g m = 100 g.

so'ma sat t r u p t g i i i a vat va mat

phang ngang la fa = 0,2. Lay g = 10 m/s^; n = 3,14. Ban dau vat nang dupe tha
nh? tai v i t r i 16 xo d a n 6 cm. Toe dp t r u n g b i n h ciia vat nang t r o n g t h o i gian ke


=> C h p n D
Cau 159. M p t tarn v a n bac qua m p t con m u o n g c6 tan so' dao d p n g rieng

bang

0,5 H z . M p t n g u 6 i d i qua t a m v a n v o i bao nhieu buoc t r o n g 12s t h i tarn van bi
r u n g len m a n h nhat?
A. 8 buoc

B. 6 buoc

C. 4 buoc

K h i b u o c chan vao t a m v a n t h i chan n g u o i da tac d u n g vao t a m van

mot

luc. Muo'n t a m v a n r u n g len m a n h nhat t h i tan so'ciia l y e tac d y n g vao t a m van

dao d p n g tren m a t phMng n a m ngang.

so m a sat g i u a vat va m|it

A. 22,93em/s

B. 25,48em/s

C. 38,22cm/s


d u o n g la chieu dan ciia 16 xo.
K h i vat chuyen d p n g theo chieu am: - k x + famg = ma = m x "
-k

r _^mg^

- m

X
\

lOOgphang

ngang la fa = 0,02. Cho gia toe t r p n g t r u o n g g = 10 m/s^. Keo vat k h o i v i t r i caf

k

= 0,02m = 2cm; co = j — = lOrad / s
Vm

X - 2 = A.cos(tot + (p) =:> V = -(oAsin(a)t + (p)

bang m p t d o a n 10 c m r o i tha nh?. Q u a n g d u o n g vat d i d u p e cho d e n k h i d i m r

Liie to = 0 -> xo = 6 c m => 4 = Aeos cf) vo = 0

han la:

=> 0 = -lOasincp => cp = 0; a = 4 em


A . 25 c m

B.25m

C. 250 cm

Huang dan gidi:

C o nang ciia vat la: W = ^ k A ^ = ^.100.0,1^ = 0 , 5 ( j )

D.250m

D . 28,66cm/s

Chpn Ox = true 16 xo, O = v j t r i ciia vat k h i 16 xo k h o n g bien dang, chieu

Huang dan gidi:

Cau 160. M p t con lac 16 xo c6 d p c i i n g 100 N / m , vat nang c6 k h o i l u p n g

tien la:

Huang dan gidi:

D . 2 buoc

p h a i bang tan so'dao d p n g rieng ciia t a m v a n .
N
G p i N la so buoc chan: => — = 0 , 5 = > N = 6 b u o c chan => C h p n B


tu thoi d i e m tha den t h 6 i d i e m vat qua v j t r i 16 xo k h o n g b i bien dang Ian dau

=> X - 2 = 4cos(10t) (cm)
K h i 16 xo k h o n g bien dang: x = 0 =:> coslOt = -1/2 = cos27i/3

t = n/15 s


Vtb =

7t/15

[jjen '^^

90
a 28,66 cm/s
3,14

ila
A. 0,05

=> Ch(?n D
Cau 163. Mot con lac 16 xo dao dong tat dan cham. Cu sau moi chu ki, bien do
dao dong cua no giam 0,5%. Nang lugng dao dong cua con lac bi mat di sau
moi dao dong toan phan la
A. 0,5%
vau

B. 1%


.ca/.n:

Ta c6:

^
A

.r

C. 1,5%

;r V. Huangddngidi:

= 0,5% => 1 - — = 0,005
A

,;,„^_

C. 0,005
Huang dan gidi:

D. 0,0025

Gpi A i , A 2 la bien dp dao dpng tai hai thoi diem each nhau nira dao dpng.

D. 2%

£)p giam CO nang cua con lac: AW = ^ k ^ A j - A j

, , , ,,,,

Congcua lire ma sat: AA^^, = F ^ , { A J

— = 0,995
A

+ A2)

= nmg(Ai +

A2)

Do AW = A A ^ , = > A j - A 2 = ^ ,!,v.

1
7 2
Nang trong chu k i tie'p theo: W = — mco A

W

B.0,025

f a tim bieu thue ciia dp giam bien dp sau moi dao dpng toan phan.

1
A2
Nang lugng trong chu k i dau: W = - mco2 A

Suy ra:

dpng toan phan va h$ so'ma sat giiia qua cau voi day kim


Dp giam bien dp trong moi dao dpng: AA = 2 ( A J - A 2 ) =

4|img )' ^ '

Theo bai ra: 200AA = AQ = 2cm
kAA

7

= (0,995^ « 0 , 9 9 = 99%

Tudo: AA = r T r = 0,01cm va n = - ^ = 0,005
200
4mg

A

Chpn C

1< Vay nang lug-ng bi mat di la 1% => Chon B

CSu 166. Mot con lac 16 xo dat tren mat phSng nghieng goe ao = 60" so voi mat

Cau 164. Mot con lac don c6 dp dai lo = 16cm dupe treo trong toa tau a ngay v i

ngang, h^ so' ma sat gii>a vat va mat phang nghieng la | i = 0,1, vat c6 khoi

tri phia tren cua true banh xe. Con lac dao dpng manh nhat khi van toe cua


lupng m = 400 (g), lay g = 10 m/s^. Cong suat can cung cap cho con lac de no

doan tau bang 15 m/s. Lay g = 10 m M 7t^ = 10, coi con tau chuyen dpng thang

dao dpng dieu h6a voi bien dp A = 5cm va tan so f = 6 Hz la:

deu, chieu dai moi thanh ray bang
A. 12 m

B. 16 m

A . P = 180(mW)
C. 18 m

C.P = 220(mW)

D . P = 240(mW)

Huang dan gidi:

D.24m

Huang dan gidi:

Sau khi cung cap bu nang lupng cho vat thi vat se dao dpng dieu hoa:
Cong cua lire ma sat tieu hao sau moi chu ky:

Chu ki cua lye cuong buc tac dung len con lac: T = —

A^„^| = H.N.4.A = n.m.g.cos(a).4.A = 0,1.0,4.cos60.4.0,05 = 0,04(j)

Nang lupng can cung cap sau moi chu ky la:

Chu ki dao dpng rieng ciia con lac: TQ = 27t J —
VS
Con lac dao dong manh nhat khi T =

B.P = 200(mW)

§iif4

27iJ— = - ==> 1 = 2n,:^.v = 12m
8

IA^,

I =0,04 0)

Cong suat can cung cap: P = ^
p=>

=> Chpn A

AW =

,1

= AW.f = 0,04.6 = 0,24(w) = 240(mW)

Chpn D


Cau 165. Mot 16 xo c6 dp cung k = 600N/m, mot dau co'dinh, dau kia gan qun

J67. Mot dao dpng tat dan c6 bien dp giam 2% sau moi chu ki. Sau 5 chu

cau nho khoi lupng m = 300g, qua cau c6 the trupt tren mot day kim loai cang

Vi CO nang dao dpng c6n lai chie'm so'phan tram so voi co nang ban dau la:

ngang trung voi true 16 xo va xuyen qua tarn qua cau. Keo qua cau ra khoi vj
tri can bang 2cm roi tha cho qua cau dao dpng. Do c6 ma sat nho, dao dpng
ch^m dan, sau 200 dao dpng thi qua cau dung l ^ i . Lay g = 10m /

. Dp giam

l-A. 17,3%

B. 28.9%

C. 73,2%

D. 81,5%

Huang dan gidi:
Theo gia thie't sau moi chu ky bien dp giam 2% tue la:

,1


Huang dan gidi:
'


100

A

Ta
T CO t = 4s = 2T => S=2.4A=2.4.4=32cm => C h p n D

C o nang dao d o n g sau m o i chu k y g i a m d i con lai Wi:

Q^aJTl:

M p t con lac d o n c6 chieu dai 121cm, dao d p n g d i e u hoa tai n o i c6 gia

toe t r o n g t r u a n g g. L a y Tt'^ = 10 . C h u k i dao d p n g cua eon lac la:

w.
w

221

= (l-0,02)

-1-0,04

A. Is

B. 0,5s

C. 2,2s


,

„ ,

D . 2s

Huong dan gidi:

2

= ( l - 0 , 0 4 ) W = 0,96W

Chu k i dao d p n g cua eon lac T = 2nJ-

Sau hai chu ky, ca nang con lai:

= 0,96Wi = {0,9ef

= 2nP^

Vg

W

V

= 2.1,1 = 2,2s
7t2


=> C h p n C
,

Sau n a m chu k y , ca nang con l a i :

Cau 172: M p t vat n h o k h o i l u p n g lOOg dao d p n g dieu hoa v o i chu k i 0,2 s va ca

W5 = ( 0 , 9 6 f W = 0,815W = 81,5%W =^ Ch(?n D

'

'

Cau 168. M o t con lac d o n g o m vat c6 k h o i i u a n g m , day treo c6 chieu doi
1 = I m . Keo con lie lech k h o i p h u o n g thang d u n g m o t goc ao = 0,1 (rad) mi

nang la 0,18 J (moc the nang tai vj t r i can bang); lay

A. 3

cm,

v

ti so d p n g nang va the nang la
B. 4

C. 2

D.l


r

Huang dan gidi:

b u o n g k h o n g v a n toe ban d a u . Con lac chiu l y c can m o i t r u o n g d g Ian coi nhu
k h o n g d o i va c6 gia t r i bang 0 , 1 % t r o n g l u g n g ciia vat. Q u a n g d u o n g ma con

=10. T a i l i d p 372

T a c o (0 = Y

= 10n(rad/s)

lie dao d p n g ke t u luc b u o n g tay cho den luc d u n g han la.
A.

4m

B. 5 m

C. 6 m

D. 7 m

2

2

Co nang W =


=> A = 0,06m = 6cm

K h i d 6 ^ = ^

= ^

Huong dan gidi:
Ap d u n g d j n h luat bao toan va chuyen hoa nang l u p n g ta c6:
Wo = W +

=

i

A«n

W,

W,

x2

=> Wo = Anwx = Fcan.Sm.nx

=> C h p n D

^

Smax


=

Cau 173: M p t vat n h o dao d p n g dieu hoa theo p h u o n g t r i n h x = A cos47tt (t t i n h

mgl"°
^ = 500.1.ao = 5 ( m ) => C h p n B
mg
u
V
/
1000

bang s). T i n h t u t=0, k h o a n g t h a i gian ngan nhat de gia toe eua vat c6 d p I a n
bang m p t nua d p Ian gia toe cue dai la
A . 0,083s.
B. 0,125s.

Cau 169: M o t con lac 16 xo c6 k h o i l u a n g vat nho la m j = 300g dao d p n g dieu
hoa v o i chu k i I s . Ne'u thay vat nho c6 k h o i l u p n g m bang vat n h o c6 khoi
l u p n g m ' t h i con lac dao d p n g v a i chu k i 0,5s. Gia t r i m2 bang
A. lOOg

B. 150g

C.25g

= 0,5Ti => 2 7 c ^ | ^ = 0,5.27i

=^ m ' = ^


Ta

CO

tai v j t r i l a i = 0 , 5

amax

t h i 1x1= 0,5.A . K h o a n g t h o i gian ngan nhat tir

'a A i = 8cm, A2 =15em va l^eh pha n h a u ^ . Dao d p n g t o n g h p p cua h a i dao
= 75gam

Chpn D

•^Png nay eo bien d p bang
A.7em.

B. 11 em.

C.17em.

Huang dan gidi:

d u a n g vat d i d u p e t r o n g 4s la:
B. 16 c m

Huong dan gidi:


£ | u 174: H a i dao d p n g deu hoa eiing p h u o n g , cung tan so' c6 bien dp Ian l u g t

Cau 1 7 0 : M o t v a t nho dao d p n g d i e u hoa v o l bien dp 4cm va chu k i 2s. Quan;-

A. 8 cm

C. 64 c m

D . 0,167s.

X = A den X = 0,5.A la t = T/6 = 0,5/6 = 1/12 = 0,083. ^ C h p n A

D . 75 g

Huang dan gidi:
Ta c6:

C. 0,104s.

D.32 c m

H a i dao d p n g v u o n g pha nen bien dp dao d p n g tong h p p :

D . 23 em.


end

A = 7 A I + A | =17cm
=>ChonC


Dt

3.

- ' '

S O N G C O

L

Cau 175: H a i con lac d o n c6 chieu dai Ian l u g t la 81 cm va 64 cm d u g c tree Q
tran m o t can p h o n g . K h i cac vat nho ciia hai con lac dang 6 v i t r i can bang

J

H E T H O N G H O A K I E N THL/C

d o n g t h o i t r u y e n cho c h i i n g cac van toe cung h u a n g sao cho hai con l i e dao

I

Song CO hpc va cac dac t r u n g ciia song

d o n g dieu hoa v o i c i i n g bien do goc, t r o n g hai m a t p h a n g song song v d i nhau

^) Dinh nghta song ca hoc

Gpi At la khoang t h o i gian ngan nha't ke t u liic t r u y e n v a n toe den liic hai day
treo song song nhau. Gia t r i At gan gia t n nao nha't sau day?

A . 8,12s.

B. 2,36s.

C. 7,20s.

>.„
D. 0,45s.

Huang dan gidi:
PT dao d p n g xi, X 2 : xi = A cos a),t
^

—
2

;

tmin

trtfong vat chat theo t h o i gian.
-

7t

2

A

/•

= COS

(0,t

2

-

C. 6 em.

D . 12 cm.

chu k i 2 s. Tai t h o i d i e m t = 0, vat d i qua can bSng O theo chieu d u a n g . Phuong

-

X =

5 cos 2 x t + ^ (cm)
2

(p =

Tan so f ciia song la tan so dao d p n g chung cua cac p h a n tCr vat chat k h i

CO song t r u y e n qua ^ = ~

D. X = 5cos

diem d o k h i song t r u y e n qua. Thong t h u o n g cang ra xa t a m tao song t h i bien


2

(cm)

= n (rad/s)

K h i t = 0 vat d i qua can bang O theo chieu d u o n g : x = 0 va v > 0
=> eoscp = 0 va sintp < 0

C h u k i T cua song la chu k i dao d p n g ehung eiia cac p h a n t u vat chat k h i

B. X = 5cos 2 7 r t - ^ (em)
2j

Huang dan gidi:
Ta CO A = 5cm; co = 2n/T= lull

Song dpc: l a song eo p h u o n g dao dpng ciia cac p h a n t u m o i t r u o n g t r i i n g

CO song t r u y e n qua va bang c h u k i dao d p n g ciia n g u o n song.
-

t r i n h dao d o n g eua vat la

C.

Song ngang: la song c6 p h u o n g dao d p n g eua eae p h a n t u m o i t r u o n g

voi p h u o n g t r u y e n song. Song dpc t r u y e n dupe t r o n g chat ran, chat l o n g va


Cau 177 : M o t vat nho dao d o n g dieu hoa doc theo true O x v o i bien d p 5 em,

(em)

.\nrn'
c) Cac dai luang dac trung cua song ca hoc

Bien d p = chieu dai q u y dao/2 = 12/2 = 6cm => C h p n C

2

.Song CO hpc k h o n g t r u y e n dupe t r o n g ehan k h o n g .

chat k h i . V i d u : song nen dan dpc theo m p t 16 xo...

Huang dan gidi:

A . X = 5cos

^

long va t r o n g chat ran. V i d u : song tren m a t nuoc,...
-

B. 24 em.

K h i song t r u y e n t r o n g m p t m o i t r u o n g t h i cac phan t u ciia m o i t r u o n g

vuong goc v o i p h u o n g t r u y e n song. Song ngang t r u y e n dupe tren be m a t chat

Dao d o n g nay c6 bien do la
3 cm.

tmt'f'

G o m hai loai la song ngang va song dpc

2

Cau 176 : M o t vat n h o dao d o n g dieu hoa theo m o t q u y dao thang dai 12 cm.
A.

;>

If) Phan loai song ca hpc

= 0,458s => Chon D

= -

Song C O hpc la cac dao d p n g dan h o i d u p e Ian t r u y e n d i t r o n g m o i

.

chi CO pha dao d p n g eua c h i i n g dupe t r u y e n d i .

cos ( 0 , t

X2 = A

H a i day song song nhau k h i xi = X2 hay cos (0,t
'
ra

, ,

chi dao d p n g q u a n h v i t r i can bang ciia c h i i n g ma k h o n g chuyen d o i theo song,

.

Suy

r,

-nil.

-

2)

(cm) => C h p n A .

;j'.)

Ciia

V a n toe t r u y e n song v la van toe truyen pha dao d p n g (khac v o i v%n toe
cac phan t u dao dpng). N o la quang d u o n g song t r u y e n d i d u p e t r o n g m p t

don vj t h o i gian. T r o n g m p t m o i t r u o n g xae djnh, v bang hang so'.
-

P h u o n g t r i n h dao d o n g la x = 5eos

Bien d p song a tai m p t diem la bien d p dao d p n g ciia p h a n t u vat chat tai

Buoe song A l a khoang each g i i i a hai d i e m gan nhau nha't tren cung m p t

p h u o n g t r u y e n song dao d p n g cung pha. Buoe song eQng la q u a n g d u o n g ma
Song t r u y e n d u p e t r o n g m p t chu k i .
-

Cong t h i i c lien h ^ T, f, v. A: >. = v.T = j ;

'
1
T =j;

1
{ =

'

A.
v = A..f = -

-

N a n g lu-ong song: Qua t r i n h t r u y e n song la qua t r i n h t r u y e n nang luong

N a n g l u o n g cua song tai m p t v i t r i chinh la nang l u g n g cua cac p h a n t u
chat dao d p n g tai d o va nang l u o n g cua song ty le t h u a n v o i b i n h p h u o n g hi^^
dp dao d p n g .

^

Giao thoa song

\ sokhdi
^

Giao thoa song la su t o n g h o p hai hay nhieu song ket h o p t r o n g k h o n g
an t r o n g do c6 n h u n g cho bien dp song tong h p p dupe tang c u o n g hoac eo

D o i v o i song phSng nang l u o n g song g i a m ty le v o i khoAng each. D o i voj
song cau nang l u o n g giam ty le v o i b i n h p h u o n g ban k i n h . Song p h a n g ne'u b6
qua m a t m a t t h i bien d p d u p e xem la k h o n g d o i nen nang l u p n g t r u y e n qug

^hiJnS

s

P h u o n g t r i n h song la bieu thuc cho phep xac d i n h l i d p u ciia t i m g phan

t u X theo t h o i gian t.

Gia s u p h u o n g t r i n h song tai O c6 dang: U Q = a.cos(a)t)

-

P h u o n g t r i n h song tai M each O m o t doan d (song t r u y e n t u O den M ) :
aeoso)

thay doi theo t h a i gian.
J,) phuong
*

= acos

" i M = aeos

= a cos CO

= acos a)t + 2 7 t —

+

H a i d a o d p n g c i m g p h a : A(]) = k 2 n (voi k e

+

H a i d a o d p n g n g u p c p h a : A(|) = (2k + 1).TT (voi k €

+

H a i d a o d p n g v u o n g p h a : Acjj = (2k + 1). ^

COt + CP] -

27rdi'

= " i M + " 2 M =2acos

= 27t

MN

Z)

Acp

d2-di

1 Acp
— +—<
1.
In
1
X

t r u y e n c u a s o n g d e u d a o d p n g t u a n h o a n theo t h a i g i a n v o i c h u k i T. V i du,

d i e m P CO tpa d p d se c 6 l i d p bien thien theo p h u o n g t r i n h Up =aeos — t - 2 7 l T
X

*

,
1 Acp
k < —+
X 2n

Acp ,
1
+ -X2' 2n
"
X

Truong

1

song, l i d p u bien t h i e n t u a n hoan theo tpa dp, xet tai t h o i d i e m to ta c6
—

2K

tn

- — X

d-i-di

X
Acp

2

v o i Acp = cp2 - cpi

(keZ)
1

Acp
+•
2 2n

(keZ)

hop hai nguon eung pha

P h u o n g t r i n h song tai hai n g u o n ciing p h u o n g Si, S2 eo dang:
Uj =acos(cot) va U 2 = aeos (cot)
-

P h u o n g t r i n h song tong h p p tai M :

T i n h chat t u a n hoan theo k h o n g gian ciia song: tren toan m i e n truyen
^27t

cos

qua chinh la so gia t r i k n g u y e n thoa m a n :

Z)

T i n h c h a t t u a n h o a n theo thoi g i a n eiia song: m p i d i e m t r e n d u o n g

= aeos

27rd2^
C0t + (p2 ~

Tren doan thSng noi \\x Si den S2 (S1S2 = 1) so cue dai, cue tieu giao thoa d i

Z)

(voi k e

' ' '

va u 2M = aeos

Bien d p dao d p n g tai M : A,^ = 2a cos

P h u o n g t r i n h tren c h o thay s o n g c6 t i n h t u a n h o a n theo t h o i g i a n va theo

+

v

Phuong t r i n h song tong h p p tai M :

khong gian.

174

Tncong hop hai nguon CO do lech pha bat ky

t, ,

Phuong t r i n h song tai M do hai n g u o n t r u y e n t o i :

(Ot- - 2 7 1 -

t r u y e n song each n g u o n O Ian l u p t la dM va dw: Acp = 27i

x,t

f

D p l^ch pha dao d p n g giira hai d i e m M va N bat k i t r o n g m o i truong

+

l '

trinh song tong hop tai M

= acos(cot + c})i) va U2 = aeos(a)t + c|)2)

P h u o n g t r i n h song tai M ' each O m p t doan d ' (song t r u y e n t u M ' den O):

-

^'^"^

p h u o n g t r i n h song tai hai n g u o n eung p h u o n g Si, S2 c6 dang:

-

-

^^"^

H a i song ket h p p la hai song do hai n g u o n ket h p p phat ra.

2. P h u o n g t r i n h s o n g

Uf^.

'^'^^ '^^

H a i n g u o n ket h p p la hai n g u o n eo eung tan so va c6 d p Ipch pha k h o n g

m p i d i e m la n h u n h a u .
-

nietn

U M = U I M + L I 2 M

= 2a cos

d2

- d i

cos

^

Bien d p dao d p n g tai M : Aj^^ = 2a cos

71

d2+di
cot-71-

d2-di


Tren doan thang noi tir S i den S2 (S1S2 = 1) so'eye dai, cue tieu giao tho,
qua ehinh la so'gia trj k nguyen thoa man:
+

So'cuc dai: - — < k < —
•

•

X

(k e Z ) ;

• rnpt eye dai giao thoa; ne'u hai n g u o n dao d p n g n g u o c pha t h i d u o n g t r u n g
trifC la r " 9 t eye tieu giao thoa.
T r o n g doan n o l g i u a hai n g u o n S1S2 khoang each g i i i a eac v a n eye d a i
hoac eye tieu l i e n tie'p bang n h a u va bang ^ ; K h o a n g each g i i i a v a n eye d ^ v a

X

+

So cue tieu: - — - —
*

Trucmg hap hai nguon ngugc

-

D o lech pha Acp = (2k + 1 )7T

— -—

(keZ)

(.yc tieu lien ke tren S1S2 la -

pha

4
.

Bien do dao d o n g tai M :

d, - d i

= 2a cos n—

7t^

Song d u n g la t r u o n g h o p dac biet eua giao thoa song: D o la s y giao thoa

giua song t o i va song phan xa ciia no tren cung m o t p h u o n g t r u y e n song.

- +—

\)) Phuang
Tren doan thang noi t u S i den S2 (S1S2 = 1) so eye dai, eye tieu giao thoa di
qua c h i n h la so'gia trj k n g u y e n thoa m a n :
+

So eye tieu: - — < k < —

ni^m

Song CO cae n i i t song va b u n g song c6' d i n h t r o n g k h o n g gian g p i la

song d u n g .
-

^

,

Songdimg

^) Mot sokhdi

P h u o n g t r i n h song tong h o p tai M :
d , - d , 71
" M = U i M + " 2 M = 2 a c o s n—^—+ - cos
X
2

•

-

trinh song tai mot diem M bat ky

Xet doan day A B = 1, d a u B co d i n h d a u A dao d p n g v o i p h u a n g t r i n h :

u = acos cot.
-

(keZ);

P h u a n g t r i n h dao d o n g tai d i e m M bat k y each d i e m p h a n xa eo d i n h B

27id^
^
. 2nd

mpt doan d : u , ^ = 2a.sm
.cos (Ot-

+

So eye d a i :

X

*
;v c-

2

(k€Z)
X

2

Trucmg hap hai nguon vuong

-

pha

+ M dao d p n g v o i bien d p eye dai hang

D Q lech pha A(p = (2k + 1) |

dj-dj

= U i K 4

IM

+u

•'2M

= 2a cos

n—

X

d, +di

n

^ + — cos

X

d, - d ,

71

X

4

Bien d p dao d p n g tai M : A j ^ = 2a cos n—

+(p,

(Pi

+ M dao d p n g v o i bien d p eye tieu bSng AMmin = 0 (nut) k h i d = k ^ ; k 6 Z

2

c) Dieu ki^t deed song ditng tren day

- +—

Tren doan thSng n o i t u S i den S2 (S1S2 = 1) so eye dai, eye tieu giao thoa t^'

+

Socuctieu:

- — + i < k < —+ - ( k e Z ) ;
X 4
X 4 ^
'

+

So'cuc d a i :

•
•

— — < k <
X 4

-

—-— f k e Z ) ,
X 4 ^
' '

-

Day dai 1 co d j n h hai d a u : 1 = k ^ (Chieu d a i cua day bang m p t so'

nguyen Ian nira buoc song)

qua c h i n h la so'gia trj k n g u y e n thoa m a n :

L u u y:

AMm« = 2a ( b u n g song) k h i

d = (2k + l ) ^ ; k e ' Z

P h u a n g t r i n h song tong h o p tai M :
U^i

. 27td

Bien d p dao d p n g tai M : A j ^ = 2a sm

•

; :>

Tgp h o p cac cue dai va eye tieu giao thoa la eac ho d u o n g hypebol

fi'i"^^

+

So b u n g : k

/

+

Sonut:k + l

'*

+

So'bo song: k

+

Q u a n h ^ g i i r a k v a i f : l = k — = k - ^ = > - ^ = -^
2

2r
1^2
n

+

' <

Buoc song dai nhat Amax = 21 k h i k = 1 (chi co 1 bo song).

xen ke n h a u . Ne'u hai n g u o n dao d p n g cung pha t h i d u o n g t r u n g t r y c eiia 51-^'
17A

177


-

D a y d a i /co d i n h rriQt dau, m p t d a u t y do: I = ( 2 k + 1 ) — (Chieu dai ,1

N e u d u n g d o n v j la Ben (B) t h i : L = l o g : ^

day bang m o t so' le Ian m p t p h a n t u b u o c song).
+

Sobung:k + l

+

S6'nut:k + 1

d) Do van toe truyeh
-

N e u d u n g d o n v i la dexiben (dB) t h i : L = l O l o g —

f'j^L-

IQ
song

^) Cac dac trtmg sinh U cua dm
.

D u n g hien t u p n g song d u n g , tit t h i n g h i e m , ta d o dup-c b u o c song >

^. p.

thanh

D p cao: La dac t r u n g sinh l i eua am, p h u thupc vao t a n so' a m . A m cao

bang each d o d u p e X/2 khoang each giua hai n u t song l i e n tie'p hoac hai byng

(tharih) la a m co tan so' am I o n ; a m thap (tram) la am eo tan so' a m n h o . D p thap

song l i e n tie'p va d o d u p e tan so' f.

|,ay cao ciia a m con d u p e h i e u qua s u t r a m hay b o n g cua am.

-

a) Song

+

dm

Song a m la song co hpc ma tai con n g u o i c6 the cam n h a n d u p e . Song am

CO tan so n a m t r o n g khoang t u 16Hz den 20.000Hz. N g u o n a m la ba't k i vat nao
phat ra song a m .
-

Cac a m co tan so tren 20 OOOHz g p i la sieu a m (tai d o i cam t h u dupe).

-

Cac a m co tan s o d u o i 16 H z gpi la ha am (tai cho cam t h u dugfc)

b) M6i trubng truyen dm vd van toe truyen
-

dm

Song a m t r u y e n d u p e t r o n g cac m o i t r u o n g vat cha't d a n h o i n h u ran,

long, k h i . Song a m k h o n g t r u y e n d u p e t r o n g ehan k h o n g . N h i r n g vat l i ^ u nhu
bong, n h u n g , n h i r n g tarn xo'p t r u y e n am kem, ehiing d u p e d u n g de l a m vot
l i f u each am.

'

-

D p to: La m p t dac t r u n g sinh l i ciia am, p h y thupc vao m i i c c u o n g d p a m

{pxbng d p a m va tan so am).

5. Song am
-

-

D u n g eong t h u c v = Xf ta t i m d u p e v a n toe v .

Gia t r i c u o n g d p am I be nha't ma tai n g u o i con cam n h a n d u p e g p i la

nguong nghe. Gia t r j cua n g u o n g nghe p h u thupc vao tan so'.
+
thi gpi
+

Gia t r j I nao d o d i i I o n l a m tai nghe co cam giac nhue n h o i , d a u d o n
n g u o n g d a u . N g u o n g d a u k h o n g p h u thupc vao tan so'.
M i e n I n a m t r o n g khoang n g u o n g nghe va n g u o n g d a u g p i la m i e n

nghe dupe. M i e n nay r p n g hep p h u thupc tan so'.
-

A m sac: La m p t dac t r u n g sinh l i ciia am, p h u thupc vao tan so' am, bien

dp song a m va cac t h a n h p h a n ca'u tao eiia am, t i i c la p h u thupc vao d o t h i dao
dpng cua a m .
+

A m CO ban va hoa am: M p t nhac cu phat ra m p t a m co tan so fo (am co

ban hay hoa a m t h u nha't) t h i bao g i o cung phat ra d o n g t h o i cac hoa am t h i i 2,
3,... CO tan s6'2fo, 3fo,...Do hien t u p n g do, am phat ra t u m p t nhac cu la s u tong

V a n to'e t r u y e n a m p h u thupc vao t i n h d a n h o i , m a t d p v a nhiet d p cua

hpp cua am co ban va cac hoa am, no dupe gpi la nhac am, t u y no co tan so'cua

m o i t r u o n g . V a n toe t r u y e n am t r o n g chat ran Ian h o n v a n to'e t r u y e n a m trong

i m CO ban fo n h u n g d u o n g bieu d i e n cua no k h o n g eon d u o n g h i n h sin dieu

chat l o n g , v a n toe t r u y e n am t r o n g chat l o n g Ion h o n v a n toe t r u y e n am trong

hoa ma la m p t d u o n g cong tuan hoan phue tap co chu k i , ta g p i n o la d o thj

chat k h i .
-

dao d p n g cua am. H p a am nao co bien d p I o n nha't se quyet d j n h d p cao cua am

Song a m t r u y e n t r o n g chat k h i la song dpc; song a m t r u y e n t r o n g chat

phat ra.

+

ran bao g o m ca song dpc va i o n g ngang.
c) Cac dac tntng vdt ly cua dm

thanh

-

T a n so am: f t u l 6 H z d e ' n 20.000Hz.

-

C u o n g d p am va m u c c u o n g dp am:
+

E
diem do. I = — =
S-t

+

^ khae n h a u phat ra sac thai am t h a n h ma ta nghe khac nhau la d o so' hpa a m
toan khac nhau.
+

C u o n g d p a m I tai m p t d i e m la nang l u p n g t r u y e n t r o n g m p t d o n v!

t h o i gian qua m p t d o n v j d i ^ n tich dat v u o n g goe v o i p h u o n g t r u y e n am to'
E

P
—

47tR^t

u n g v o i tan so' f = 1 OOOHz).

' ^ o n g t u a n hoan.
^guon
-

47iR^

g i i i a c u o n g d p a m I tai d i e m d a n g xet va c u o n g d p am chuan lo (lo = 10"^^ W / m

A m d u p e phat ra t u m p t tieng no, tieng go vao t a m k i m loai,.. g p i la

am, chving k h o n g co tan so' xae d j n h , d o t h i cua ehung la d u o n g cong

. D o n v j : W / m ^ (oat tren met v u o n g )

M u c c u o n g d p am L la dai l u p n g do bang loga thap p h a n cua ty s^'

Dao d p n g am tong h p p co t i n h tuan hoan, d a n g cang p h u e tap t h i so

hpa am cang n h i e u , am sac cang p h o n g p h i i . C i m g m p t am sol d o n h i e u d u n g

^

nhac dm vd hgp cong

huong

N g u o n nhac a m la n h i r n g n g u o n phat ra nhac a m , tuc la phat ra n h u n g
CO tan so xae d j n h , m o i nhac cu la m p t n g u o n nhac am.

-

H o p cong h u o n g la m p t vat rong co kha nang epng h u o n g d o i v o i nhieu
so khac n h a u v . tang c u o n g n h i r n g am co cac tan so d o .


X
^n toe truyen song la: ^ ^ j

II. C A C D A N G B A I T A P V A V I D y MINH HQA
Dong^: XAC D I N H C A C D A C T R l / N G CUA S O N G C O H Q C
* Phuong phap giai: - Su dung cac cong thuc: to = 27tf =

^= v T =

Limy:
. O day hoe sinh c6 the nham mpt trong cac truong hpp nhu sau:
+

Vi

50
^ ^ 50cm / s =5> Chpn D

1: Mot nguoi quan sat mpt chiec phao tren mat bien thay no nho cao

1-:'

:

Khoang each 2 ngpn song ke nhau la 0,5m nen A = I m = 100cm
Van toe truyen song la: v = ^ =

= lOOem / s => chpn B

len 6 Ian trong 10 giay va thay khoang each giiia hai ngpn song ke nhau la
+

0,2m. Van toe truyen song bien bang:
A. 12cm/s

B. lOcm/s

C. 24cm/s

Co 10 ngpn song di qua truoc mat trong 9s nen chu k i song la: lOT = 9s
^ T = 0,9s

D. 20cm/s

Huong dan giai:

=>Van toe truyen song la: v = ^ =

= 55,56cm / s => chpn A

Khoang thoi gian giiia 6 Ian nho lien tiep la 5 chu ki => 5T = 10 => T = 2s
Khoang each 2 ngpn song ke nhau la 0,2m => A = 0,2m = 20cm

+

n
•

A, 20
Van toe truyen song la: ^ ^ ^ ^ ^ 10cm / s

Co 10 ngpn song di qua truoc mat trong 9s nen chu k i song la: lOT = 9s

^

z^ChpnB

=> T = 0,9s. Khi do van toe truyen song la: v = — = - — = 111,11cm / s
T 0,9
=> chpn C

Lira y:
-

O day hoc sinh c6 the nham nhu sau:

+ Khoang thoi gian giOa 6 Ian nho lien tiep la 6 chu ki => 6T = 10 => T = 5/3s

"k 20
Van toe truyen song la: v = — = — = 12cm / s => ehpn A

"

Dang 2: B A I TAP V E P H l / O N G TRINH SONG V A D O L E C H PHA
G I 0 A HAI DIEM TREN PHLfONG TRUYEN SONG
* Phuong phap giai

3

+

Khoang each 2 ngpn song ke nhau la 0,5m nen A = I m = 100cm

Khoang each 2 ngpn song ke nhau la 0,2m => A = 0,4m = 40cm
-> Van toe truyen song la: ^ = ^ = ^

^

-

Gia su phuong trinh song tai O c6 dang: Uy = a.cos(a)t)

-

Phuong trinh song tai M each O mpt doan d (song truyen tir O den M ) :

IQcm / s => ehpn D
u ^ = acosco t — = aeos cot - 2 7 1 -

+

Khoang thoi gian giCra 6 Ian nho lien tiep la 6 chu ki => 6T = 10 => T = 5/3s
Khoang each 2 ngpn song ke nhau la 0,2m
Van toe truyen song la: v = — = ^

A = 0,4m = 40cm

-

Phuong trinh song tai M each O mpt doan d (song truyen tit M den O):

= 24cm / s => ehpn C

acosco

3

-

(

t +

d^
-

= aeos

COt +

27t-

Dp lech pha dao dpng giira hai diem M va N bat ki trong moi truong

V i du 2: Mpt nguoi quan sat song tren mat ho thay khoang each giua hm
ngpn song lien tiep la 0,5m va c6 10 ngpn song di qua truoc mat trong 9^
Van toe truyen song tren mat nuoe bang:
A. 55,56cm/s

B. lOOcm/s

C lll,llcm/s

D. 50em/s

t''uyen song each nguon O Ian lupt la dM va dw: Acp = 27t+

Hai dao dpng cung pha: Acp = k27T (k e Z)

+

Hai dao dpng ngupe pha: Acp = (2k + 1 ).TT (k € Z)

Huong dan giai:
Khoang each 2 ngpn song ke nhau la 0,5m nen A = 0,5m = 50cm
Co 10 ngpn song di qua truoc mat trong 9s nen chu k i song la:
9T = 9 s ^ T = ls

+ Hai dao dpng vuong pha: Acp = (2k + 1). ^ (k e Z)


*

Cdc vi du tninh hpa:

Huong dan gidi:

V i d^ 1: Song t r u y e n tai m a t chat long v o i buac song 0,1m. P h u a n g trhih

27cd

(cm)

.

P h u a n g t r i n h song t?ii M c6 dang: U j ^ = a cos cot--

.

Thay U j ^ = 2cm, t = T/2 va d = A/3 vao p h u a n g t r i n h t r e n ta dugic:

dao d o n g t a i n g u o n O c6 d a n g UQ = Scoscot ( m m ) . P h u o n g t r i n h dao d p n g tai
d i e m M each n g u o n O m p t doan 6cm theo p h u a n g t r u y e n song la:
A . u ^ = 5cos(cjt + l , 2 7 t ) ( m m )

B. u , ^ = 5cos(cot + 1 2 0 7 i ) ( m m )

C. U [ ^ = 5cos(a)t - 1207T) ( m m )

D . u ^ = 5cos(ajt - l,2n) ( m m )

I

^;

= 5 cos cot
V

271.6^

J

10

=> C h o n D .

=> 2 = acos ( r t - — ) = — = > a = 4 c m => C h p n B.
3
2

A. A. = Tia

B. X. = 27ia

C. A. = 2na/3

D . A, = 37ia/2

Huong dan gidi:
In

j^^y

6 d a y hpc sinh c6 the n h a m n h u sau:
+

3.x

d i i n g la:
= 5cos(cot - l , 2 7 t ) ( m m )

Ltfu y:
-

2

eye d a i eiia p h a n ttr m o i t r u o n g bang 2 Ian toe d p t r u y e n song, b i e u t h i i c

P h u a n g t r i n h song tai M d o O t r u y e n den c6 dang:
2nd'

T

V i d u 3: M o t song co hpc ca bien d o a, b u o c song A. Biet v a n toe d a o d g n g

Huong dan gidi:

(

u ^ = 5 cos cot

2 = acos

N h a m ve h u o n g t r u y e n song:

V i d u 4: N g u o n song co t r u y e n tren m p t spi d a y d u a c m o ta b a i p h u o n g

P h u o n g t r i n h song tai M do O t r u y e n d e n c6 dang:
U j ^ =5cos| cot +

2nd

.a
J ^ V d d m a x J ^ ^ T ^ J _ . T = na ^ C h o n A .
^
2
2
2

= 5 cos cot +

27C.6

10

trinh u = acos(27rt - 0 , l 7 t x ) , t r o n g d o u va x d u p e d o b a n g cm, t d o b a n g s.

= 5cos(cot + l , 2 7 i ) ( m m )

Tai m o t t h o i d i e m d a cho d p lech p h a dao d p n g ciia h a i p h a n t u t r e n day
each n h a u 2,5 c m l a

=> chpn A .
+

B.^

A.^

N h a m ve d o n v i chua d o i t h o n g nhat:
P h u a n g t r i n h song tai M d o O t r u y e n den c6 dang:
Uj^

(

.

5cos cot +

27rd^

271.6^

= 5 cos cot + -

10

= 5 cos (cot + 1207t) ( m m )

N h a m ca h u o n g t r u y e n va d o n v i chua d o i :
P h u a n g t r i n h song tai M d o O t r u y e n den c6 dang:
UM=5COS

r cot

2nd^

= 5 cos cotV

27t.6^

0,1

J

-

P h u a n g t r i n h song tai m o t d i e m M co dang: u = a cos c o t - -

-

Theo gia t h i e t u = a cos (27it - 0, ITIX)
nen

= 5cos(cot-1207t) ( m m )

27rx
u = a c o s ( 2 7 i t - 0 , l 7 t x ) = acos cot--

V i d\ 2: N g u o n song co t r u y e n theo m o t d u a n g thang, n g u o n dao d p n g v o i
p h u a n g t r i n h UQ = acoscjt (cm). M p t d i e m M tren p h u a n g t r u y e n song each
= 2cm. C o i b i e n d p song

B. 4em

0 , l 7 i x = — ^ A . = 20em

27t.2,5 7t,
20

4^

^

Chpn D.
V i d u 5: M o t song t r u y e n tren m a t nuoe bien co b u a c song X = 2 m . K h o a n g
each giira h a i d i e m g a n n h a u nhat tren c i m g m p t p h u a n g t r u y e n song dao

k h o n g b j suy giam, b i e n d p song a la:
A. 2cm

27tX

D p l$eh pha dao d p n g ciia hai phan t u tren day each n h a u 2,5 c m la:
2iid
Acp = . ^

chon C.

n g u o n d = A/3 t a i t h o i d i e m t = T/2 c6 l i d p

4

Huang dan gidi:

=> c h p n B.
+

8

6

C. 2 x / 2 e m

D. 4 V 2 c m

A. 1,25 m

B.2,5m

C. 5 m

D. 1 m


Huong dan gidi:

phuong trinh song tai M do hai nguon truyen tai:

Khoang each giua hai diem gan nhau nhat tren ciing mpt phuong truy^Y,

MM = acos cot + (pj -

song dao dpng ngupe pha nhau la: d = X/2 = 1 m. => Chpn D
V i dv 6: Mpt song truyen tren mat nuoc bien c6 buoc song X = 2m. Khoang

27id, ^

C . 0,5 m.

D . Mpt gia tri khac.

Acp
+ — cos

d-, - d i

n—

UM="iM+"2M=2acos

dong l^ch pha nhau n/2 la:
B. 3 m.

-

27td2^

phuong trinh song tong hpp tai M :

each giiia hai diem gan nhau nhat tren cung mpt phuong truyen song dao
A . 0,75 m.

va U j M =aeos cot +

d, - d i

Bien dp dao dpng tai M : A,^ = 2a cos

•K—

- +

Acp
—

d, +di

(D, + ( p ,

X

1

voi Acp = cp2 - cpi

Huomg dan gidi:
Tren doan thang noi tu Si den S2 (S1S2 = 1) so cue dai, cue tieu giao thoa di

Dp l?ch pha dao dpng ciia hai phan tu tren phuong truyen song each nhau
d l a : A(p = — ( r a d )
X

qua chinh la so gia trj k nguyen thoa man:

V . * M J^HW*.**

—

Theo gia thiet A(p = — =>
^
2

1

;i

= — => d = — = 0 , 5 ( m )
2
4
^ ^

1

Acp ,
1 Acp
+ — < k < —+ —
27t

X
In
1

=> Chpn C
2

V i d u 7: Song truyen tren day voi van toe 4 m/s c6 tan so' song thay doi tu 2 2
pha voi nguon. Buoc song truyen tren day la:
B. 1,6 cm.

C . 16em.

1

271

1

Acp
+—

2

/,
(keZ)

27t

^

If " i t / !

. 0

«

'

Truang hop hai nguon cung pha:

H z den 26 H z . Diem M each nguon mpt doan 28 cm luon dao dpng vuong
A. 160 em.

Acp ,
+—
t mii

/,
„^
(keZj

D . 100 cm.

-

Dp l^ch pha Acp = 2kTt

-

Phuong trinh song tong hpp tai M :
d2-di

= " MM
1 M + " 2 M = 2a COS

Huong dan gidi:

COS

cp. +cp,

d, +di

[fii. •

D p l^ch pha dao d p n g ciia hai phan tu tren phuong truyen song each nhau
dla:

Bien dp dao dpng tai M : A|y^ = 2a COS

A(p = ^ ^ = ^ ^ . f ( r a d )
X
V
^
'

Tren doan thang noi tu Si den S2 (S1S2 = 1) so'cue dai, cue tieu giao thoa di

Theo gia thiet
n ,
27id, n /
^ { k + 0,5).y
(k + 0,5).400 (k + 0,5).50
Acp = - + krc =>
f = - + k7i => f =
— =
=^
-^
—
^ 2
V
2
2d
2.28
7
M a 2 2 H z < f <26Hz=> 22 H z <
V i k nguyen nen k = 3

(k + 0,5).50
' - — < 26 H z
7

2,58 < k < 3,14;

f = 25Hz.
V

Vay buoc song truyen tren day la: X- — =

400

= 16cm => Chpn C

Dang 3: B A I T A P V E G I A O T H O A S O N G C O H Q C
*

qua chinh la so'gia trj k nguyen thoa man:
+

So'cue dai: — < k < ~
•
•
X
X

i
§

+ So cue tieu:

^

-

(k

- - - - < k < - - -

^ 2

^

2

6

Z);

(k e Z)
^
^

>

'

••

-

"

Trong doan noi S1S2 khoang each giiia cac van cue dai hoae cue tieu lien

tiep bang nhau va bang ^ .

Chuy:

P h u a n g phap giai:

-

K h i giai bai toan giao thoa, can kiem tra dp l§ch pha eua hai nguon.

Phuong trinh song tong hop tai diem M:

-

K h i hai nguon dao dpng voi bien dp khac nhau thi phuong trinh song

Phuong trinh song tai hai nguon cung phuong Si, S2 c6 dang:
U j = aeos(cut + cpi) va U j = acos(a)t + 92)

tong hop tai M bat ky dupe xae djnh nhu tong hpp hai dao dpng (hai dao dpng
*hanh phan la hai song do hai nguon truyen den M) va bien dp dao dpng tai M


duQC ti'nh:

= yja^ +al+

y f d y 2: T r o n g m p t t h i n g h i e m giao thoa song tren m a t nuoc, hai n g u o n ket

la^aj cos(A(p) t r o n g d o A(p l a d p l^ch p h a ciia haj

Pi^p S i va S2 dao d o n g v o i tan so' 16 H z , c u n g pha. Tai d i e m M each hai

song t h a n h p h a n tai M .

-

pguon Ian l u g t la d i = 30 c m va d2 = 25,5 em, song eo bien d g cue d a i . G i i i a

V o i bai toan t i m so d u a n g dao d p n g v o i bien d o cue d a i , cue tieu giijg

hai d i e m M , N bat k y each hai n g u o n Ian l u g t la d j ^ , 62^,

va d u o n g t r u n g true ciia S1S2 c6 t h e m m o t g g n l o i nua. V a n toe t r u y e n

d j ^ j , d j ^ t h i lam

song tren m a t n u o c la

n h u sau:

!

A^24 em/s

+

Dat A d ^ j = d 2 M - d , M ; ^ d N = d 2 N - d i N ; Gia s u Ad^^ > Ad^j

+

N e u hai n g u o n dao d o n g eung p h a :

* '
+

*

Cue d a i : A d ^ < k > i < A d M

*

Cue t i e u : Ad^J < ( k + 0 , 5 ) ? L < A d M

C. 72 m/s

D . 7,1 cm/s

Huang dan gidi:
.

H a i n g u o n dao d o n g cung pha, tai M song co bien d p eye d ^ i nen

^ , , d 2 = k.A.
.
*

„^:^P--

G i i i a M va d u o n g t r u n g true ciia S1S2 c6 t h e m m g t g g n l o i nua nen k = 2.

V?y 30 - 25,5 = 2.A

N e u hai n g u o n dao d p n g nguge p h a :

.

*

C y c d a i : Adf^ < ( k + 0 , 5 ) > i < A d | ^

=> Van toe t r u y e n song tren mat nuoc la v = A.f = 2,25.16 = 36 (em/s) => Chgn B

*

Cue tieu: Ad^j < k A . < A d ^

Buoc song A = 2,25 cm

L u u y : Co the n h a m n h u sau
-

S o g i a t r i n g u y e n ciia k thoa m a n cac bieu thue tren la so d u a n g can t i m .

*

B. 36 cm/s

Buoc song A = 1,5 em ^ v = A.f = 1,5.16 = 24 (cm/s) =>Chgn A
Vi

Cdc vi du mirth hoa:

k = 3. Vay 30 - 25,5 = 3.A;

3: O be mat mgt chat long c6 hai nguon phat song ket hgp S i va S2

each nhau 20cm. H a i nguon nay dao dgng theo phuong thSng dung c6

V i d v 1 : T r o n g t h i n g h i e m ve h i ^ n t u g n g giao thoa song, n g u o i ta tao tren

phuong trinh Ian lugt la u^ = 5eos (407tt + n/6) (mm) va U j = 5cos(407tt + 77t/6)

m a t n u o c hai n g u o n A va B dao d o n g cung p h u o n g t r i n h : u ^ = 5cosl07tt

(mm). Toe do truyen song tren mat chat long la 80 cm/s. So diem dao dgng

(cm) va U g = 5cos(107rt + 71) (cm), v a n toe t r u y e n song tren m a t nude la

voi bien do cue dai tren doan th^ng S1S2 la
A. 11.

20cm/s. D i e m M tren mat nuoc c6 M A = 7,2 cm, M B = 8,2 c m c6 p h u o n g

B. 9.

A . u ^ =5N/2cos(207tt-7,77t)cm

B. u ^ = 5 > ^ c o s ( l 0 7 t t + 3 , 3 5 j i ) c m

C. U[^ = 1 0 V 2 c o s ( l 0 7 i t - 3 , 3 5 7 i ) c m

D . u^^ = 5 N / 2 c o s ( l 0 7 t t - 3 , 3 5 7 t ) c m

'

1
V
80 ^
D u o e song X = — - — = 4cm
^
f
20

R

So'cue dai giao thoa tren S1S2
- i < k< ^
• ^
X
2
X
V i k 6 Z nen k = - 5 , - 4 , - 3 , - 2 , - 1 , 0 , 1 , 2, 3 , 4.

Huang dan gidi:
-

Buoc song A = v/f = 20/5 = 4 (cm)

-

P h u o n g t r i n h song tong h g p tai M :

"M="iM+U2M=2acos

r:> C h p n D

1 O/l

8,2-7,2

D . 8.

Huang dan gidi:

t r i n h dao d o n g la :

u^^ =2.5. cos

CIO.

_>, - 5 , 5 < k < 4,5
,^

Co 10 gia trj ciia k, vay c6 10 d i e m tren doan S1S2 dao d g n g v o i bien d g cue
^=>ChonC.

cos (ot-n—

cos

- i
I

4

dj+dj
(pi+92
- +—
~
X
2
2

= 5V2cos(l07tt-3,357t)

d y 4: H a i n g u o n song co S i va S2 each n h a u 42 c m dao d g n g theo p h u o n g
trinh: U j = U 2 = 4cos507rt (cm) Ian t r u y e n v o i toe d g v = l,Om/s. So d i e m
' ^ o n g dao d g n g tren doan thang no'i S1S2 la ( k h o n g ke hai n g u o n ) :
_

A . 10

B. 12

C. 16

Huang dan gidi:
^.R. ,
,
,
v
100
,
1 Duoe sone X = — =
= 4cm

*
^
f
25

D.20


S S
1
S S
1
So cvrc tieu giao thoa tren S1S2 — L i . _ - < k <
- - -> - I K k < 10

.

27cd

Bien d p dao d g n g tai M : A ^ = 2a sm-

V i k e Z n e n k = -10,-9, ...,9
Co 20 gia trj ciia k, vay c6 20 d i e m tren doan S1S2 k h o n g dao d p n g -> chpn D
V i dv 5: Tai hai d i e m A va B tren mat n u o c c6 2 n g u o n song dao d o n g cung
pha, bien d p Ian l u p t la 4cm va 2cm, buoc song la 10cm. D i e m M t r e n mat
B. 4cm

C. 6cm

D. 8cm

Huang dan gidi:

«'.ln

+ ^2 + ^^\^2

+

AMmm = 0 k h i d = k — ;

2

.

Day c o ' d j n h h a i d a u : 1 = k ^
+

COS(A(P)

27r(d,-d,)
V o i Acp = — —
— = n . Suy ra A M = 2 cm. ^ C h p n A
V i dy 6: H a i d a u M va N ciia m p t mau day thep nho h i n h chi> U dupe dat
cham vao m a t nuoc. Cho mau day thep dao d p n g d i e u hoa theo phuong
v u o n g goc v o i m a t nuoc. Bie't M N = 6,5cm dao d p n g v o i tan so' f = 80Hz ; toe

+

So'nut: k + 1

+

So'bo song: k

+

Q u a n h# giira k v o i f: 1 = k — = k —
^
2
2f

+
-

So'byng: k

S6'bung:k + 1

t r o n Ion bao ca h a i n g u o n song vao trong. Tren v o n g t r o n ay c6 bao nhieu

+

Sonut:k + l

C.64

D. 66

Huong dan gidi:

Buoc song: A = v/f = 32/ 80 = 0,4cm

Khoang each giiia 2 n u t song hay hai b u n g song bat ky: d - k - ^ , k = 1,2,3...

-

K h o a n g each giOa 1 n i i t song v o i 1 b u n g song bat ky:
d - ( 2 k + l ) ^ , k = 0,1,2,3...

^,

X

M N dao d p n g v a i bien dp cue dai. K h i d o v o n g t r o n bao q u a n h hai n g u o n

-

va N cat cac van cue dai 6 66 d i e m => C h p n D.
Dang 4:

..A ,

-

< k < - => -16,25 < k < 16,25 => c6 33 d i e m tn n
X

£2

Luu y:

B. 32

So cue d a i tren M N la:

k2

Day CO d j n h m p t dau, m p t d a u t u do: 1 = (2k + 1 ) ^
+

d i e m CO bien d p dao d p n g cue dai.

.

Buoc song dai nha't Amnx = 21 k h i k = 1 (chi co 1 bo song).

d p t r u y e n song v = 32cm/s ; bien dp song k h o n g d o i A = 0,5cm. Ve m p t vong

A.33

keZ

jv M fcu

H a i n g u o n dao d p n g c u n g pha nen bien dp dao d p n g tai M :
= V^i

AMmax =

Dieu kien deed song ditng tren day

n u o c each A 25cm va each B 30cm se dao d p n g v o i bien d p la
A. 2cm

2 a k h i d = (2k + l ) - ; k € Z
V
14

+

•[ .A

T h o i gian hai Ian day d u o i thSng lien tie'p: At = •
Be r p n g cua m p t b u n g song la: 4a

M p t spi day thep cang th^ng, dat gan m p t d a u n a m cham dien thang,

BAI TAP V E SONG Dl/NG

d o n g dien qua n a m cham co tan so'f t h i day se dao d p n g v o i tan so 2f.
*

Phuong phdp:

-

M p t spi day no'i v o i n g u o n d i f n xoay chieu tan so f, day dat t r o n g

Phuong trinh song tai mot diem M bat ky:

khoang giira hai ban cua m p t n a m cham h i n h chCr U t h i day se dao d p n g v o i

-

tan so cung la f.

Xet d o a n day A B = 1, dau B co d i n h dau A dao d p n g v o i p h u o n g t r i n h :
u = acosu't.

-

P h u o n g t r i n h dao d p n g tai d i e m M bat k y each d i e m p h a n xa co' djnh 1'

mptdoand:

o
• 27id
( ^
Uj^=2a.sm
.cos cot
X
y

2nt]
X )

y

-

Tan so d o day d a n phat ra (hai d a u day c o d i n h ) : f = k — ; k e N

-

Tan so d o 6'ng sao phat ra (mpt dau bjt k i n , m p t dau de h o

^ut song, m p t d a u la b u n g song): f = (2k + 1 ) - ^ ; k e N

m p t d a u la


iv> inuuigmi

ruiunn

B I I I \ L I LJ., lup i - i \ g u y e n jjuang

irmn

i.ac

Huang dan gidi:
Dieu kien de c6 song dirng tren day trong truong hop hai dau co'djnh:
X
, 21
1 = n — => A = —
2
n
21 21
212

Voi 3 mui t a c 6 n = 3=> X = — = — = — = 0,8m => chpn B.
n
3
3
V i d u 2 : Mot day dan c6 chieu dai 90cm, khi gay phat ra am co ban tvrong
ung CO tan so f. Muo'n cho day nay phat ra am co ban f = l,2f thi phai b a n i
phim cho day ngan lai con chieu dai 1' bang
A. 60 cm
B. 75cm
C. 50cm
D. 65cm
Huang dan gidi:
dinh

I = n . ^ . A m CO ban do day dan phat ra ling voi n = 1 => 1 = ^ = ^ ( * ) i
(**)
'
Chon B

V i du 3: Song dung tren day AB co hai dau co dinh, chieu dai 40cm. Tan so

l = k.^ = k ^
2
2f
Vi 1 va V khong doi nen khi so bo song it nha't (tuc n = 1) thi tan so'f se nho
V

nha't

B. 10 bung, 10 niit

C. 5 niit, 5 bung

D. 9 bung, 10 nut

fmin = ^ •

+ Hai tan so gan nhau nha't la 200Hz va 300Hz se ung voi so' bo song lien
tiep la k va (k + 1), tiic la 1 = k

^
va 1 = (k +1)—-—. T u hai phuong trinh
2.200
2.300

tAy se CO 1 = 200
+ Thay vao bieu thuc fmin ta dugc fmin = lOOHz. -> Chgn A.
Vi d u 5: Mot day AB hai dau co dinh. Khi day rung vol tan so'f thi tren day
CO 4 bo song. Khi tan so tang them 10 Hz thi tren day co 5 bo song, van toe
truyen song tren day la 10 m/s. Chieu dai va tan so rung cua day la:
A. 1 = 50 cm, f = 40 Hz.
B. / = 40 cm, f = 50 Hz.
C./ = 5cm, f = 50Hz.
D. / = 50 cm, f = 50 Hz.
-

dao dgng cua day la 50Hz, van toe truyen song tren day la 4m/s. Tren day co:
A. 10 bung, 11 nut

KHattg^rtfT

V

Dieu kien de co song dung tren day trong truong hop hai dau co

r f
It
If
1
90
T u H v a r ) s u y r a i - = 1 =:> j ' = _ = _
= _ ^ _ = 75cm
^'
^ ' y
I f
f
i,2f 1,2
1,2

UVVH

y i du 4: Nguoi ta t^o song dirng tren mot soi day cang ngang giira hai diem
CO djnh. Song dung dugc tao ra tren day Ian lugt voi hai tan so' gan nhau
jiha't la 200Hz va 300Hz. Tan so' kich thi'ch nho nha't ma van tao ra song
dung tren day la :
A. lOOHz
B. 50Hz
C. 200Hz
D. 150Hz
Huang dan gidi:

- Dieu ki^n de co song dung tren day trong truong hgp hai dau co djnh:

Cdc vi du minh hga:

Tuong t u de co tan so' f thi ' ~ ^ ~ ^

miv

Huang dan gidi:
V
4
Buoc song: X = v.T = — = — = 0,08m = 8cm
^
f
50
Dieu ki^n de co song dij-ng tren day trong truong hgp hai dau co'djnh:

l = k.A
2
21 2 40
So bung song: k = — = —— = 10 => so byngsong la 10, so niit song la 11.
X
8
^ Chon A.

Huang dan gidi:
Day AB hai dau co'djnh nen chieu dai cua day la

• i

l = k . ^ = k . ^ = ^ f = k.-^2
2f
21
-

Khi day rung voi tan so f thi tren day co 4 bo song: f = 4 . ^

(1)

-

Khi tan so tang them 10 Hz thi tren day co 5 bo song: f +10 = 5.-^ (2)

Tu (1) va (2) suy ra ^ = 10 ^ 1 = 0,5m - 50cm .
JThay vao (1) suy ra f = 40 Hz =:> Chgn A
du 6; Mot sgi day M N treo lo lung, dau tren M gan vao am thoa dao
dgng voi chu ki T = 0,02s, toe do Ian truyen song la 2m/s. De tren day co
^ n g dung, voi 13 niit va 13 bung (ke ca hai dau), thi sgi day phai co chieu
bang bao nhieu?
- . _ A . 20 cm
B. 30 cm
C. 50 cm
D. 25 cm


Huang dan

gidi:

-

Buac song: A, = v T = 2.0,02 = 0,04m = 4cm

-

D i e u k i ^ n de c6 song d u n g tren day: 1 = k | + ^

Thay>. = 4 c m v a o n t a c 6 :

l = 2k + l

V i d u 3: M p t cai gieng sau 20m t i n h t u thanh den mat nuac. M p t em be
d u n g gain thanh gieng, d o so y da d a n h r o i ehie'c bat. Biet v | n toe t r u y e n am
(*) v o i k e N

trong k h o n g k h i la 320m/s, lay g = lOm/s^. T i n h t h o i gian t u liic d a n h r o i den
luc nghe thay tieng bat cham vao nuoc.

(**)

A . 2,4s.

-> De CO song d i r n g tren day v o i 13 b u n g va 13 n i i t t h i k = 12.
K h i d o tir (**) ta c6: / = 2k + 1 = 2.12 + 1 = 25(cm). - » C h p n D .
Dang

-

Cac bai toan t h u o n g gap o dang nay la xac d i n h cac dai lug-ng dac trung

phdp:

am ...

gidi:

T h o i gian t u k h i ehie'c bat r o i den k h i cham vao m§t n u o c la

-

T h o i g i a n t u liic bat cham vao mat nuoc den k h i nghe thay tieng bat
20

= 0,0625s
320
V a y tong t h o i gian la t = t j + t j = 2,0625s -> C h p n D .
V

-

Cac cong thuc t h u o n g su d u n g la:

Vi

D . 2,0625s.

-

h
cham vao mat n u o c la t , = — =

cho song am n h u van toe am, m u c c u o n g dp am, tan so a m , tan so cac hoa

-

"

5: B A I T A P V E S O N G A M

Phttang

-

C. 2s.

Huang dan

.

*

-

B. 2,1s.

V i du 4: M p t n g u o i d u n g biia go nh? vao d u o n g sat va each d o 1376m, eo

V a n toe: v = — = k.f; v = T
t
E E

P
C u o n g d g am: I = —- =
r- =
r
S.t
4:iR2t
47IR2
I
M u c c u o n g d o a m : L ( B ) = l o g — hay

mpt n g u o i ap tai vao d u o n g sat t h i nghe thay tieng go som h o n 3,3s so v o i
tieng go nghe t r o n g k h o n g k h i . V a n toe am t r o n g k h o n g k h i la 320m/s. Van
toe am t r o n g sat la
A . 1238m/s.

I
L ( d B ) = 101og —
^0
1: M o t song am d u p e phat ra t u m o t n g u o n a m c6 d a n g h i n h cau. Biet

Huang dan
-

flic go.

ra dupe bao toan. Xac d j n h cong suat phat am ciia n g u o n .

*

B.4,5W

C.4W

Huong dan

gidi:

N a n g l u p n g dupe phan bo deu tren dien tieh mat song la mat eau: S = 47:R
C o n g suat phat am la: P

IS = 0,06.4TT.2^ « 3 W -> C h p n D .

V i du 2; M p t la thep m o n g , m p t dau eo d j n h , dau eon lai d u p e k i c h thich de
dao d p n g v o i ehu k i k h o n g d o i va bang 0,08s. A m d o la thep phat ra la
A . sieu a m .

B. ha am.

C. nhae am.

D . am ma tai n g u o i nghe dupe.
Huang dan

C h p n B.

D . 1376m/s.

gidi:

V o i v J , V2 la v a n toe am t r u y e n t r o n g sat va t r o n g k h o n g k h i

Ta c6: V j t - V2 ( t + 3,3) = 1376 ^

t = I s ; v i = 1376m/s.

Chpn D.

V i d^ 5: M p t n g u o i d u n g gan 6 chan n i i i h i i len m p t tieng. Sau 8s nghe tieng
m i n h v o n g lai, biet van toe am trong k h o n g k h i la 340m/s. K h o a n g each t u
chan n i i i den n g u o i d o la
_

A . 1333m.

B. 1386m.
Huang dan

C. 1360m.

D . 1320m.

gidi:

Gpi / la k h o a n g each t u n g u o i d o den chan n i i i .
Theo de ra v t

gidi:

12,5HZ
Tan so cua am phat ra la f = ^1 = 12,5Hz

T
A m do la thep phat ra la ha a m

-

D.3W

C. 1336m/s.

G p i t la t h o i gian n g u o i kia nghe dupe tieng go t r o n g d u o n g sat ke t u

tai d i e m each n g u o n 2 m , c u o n g do a m la 0,06W/m2. c i a s u nang l u g n g phat
A.5W

B. 1356m/s.

21 => 1 = y = 1360m => C h p n C

V i du 6: H a i a m eo m u c c u o n g d p am h o n k e m nha 40dB. T y so c u o n g dp
9ni cua c h i i n g la:
_

A . 400

B.100.

C. 40

D . 10000

Huang dan gidi:

'

T a c6: L = l O l o g — v 6 i I i t h i L j = l O l o g ^
^0

Huang dan
Cac dap an B, C, D deu d i i n g .

(*)

^0

gidi:

Chon A

Ket l u a n nao sau day khong diing ve s u t r u y e n song co

v o i h t h i L2 =101og^

(**)

A. Song n g a n g la song m a p h u o n g dao d p n g ciia p h a n t u m o i t r u o n g

^

v u o n g goc v o i p h u o n g t r u y e n song.

T u n va D

B. Song dpe la song m a p h u o n g dao d p n g ciia p h a n t u m o i t r u o n g t r u n g

- ) . L2 - L i = l O l o g ^ - lOlogiL = l O l o g ^
^0

v o i p h u a n g t r u y e n song.

,i

^1

C. Song t r u y e n t r o n g m o i t r u o n g ran l u o n la song dpe.
V i L 2 - L i = 40 -> 40 = l O l o g - ^
^1

l o g ^ = 4 =^ ^ = l O " = 10000 ^ C h p n D.

D. Song t r u y e n t r o n g m o i t r u o n g k h i l u o n la song dpe.

h

^1

Huang dan

V i dy 7: H a i hoa am lien tiep do mq>i day d a n phat ra c6 tan so h o n kem

T r o n g m o i t r u o n g ran g o m ca song dpe va song ngang. -> C h p n C .

nhau 56Hz. H o a a m t h i i 4 c6 tan so:
A.28HZ

B.224HZ

C. 86Hz

C g u S : V a n toe t r u y e n song p h u thupe vao yeu to nao?

D. 168Hz

Huang dan gidi:
Theo bai ra: n f - (n - l ) f = 56 - » tan so a m co ban f = 56Hz

A. N a n g l u p n g song.

B. T a n so dao d p n g .

C. M o i t r u o n g t r u y e n song.

D. Buoc song.
Huang dan

Vay tan so hoa a m t h u t u f4 = 4f = 4.56 =224 H z . -> C h o n B.

gidi:

Van toe t r u y e n song p h y thupc vao m o i t r u o n g t r u y e n song. - > C h p n C

III. B A I T A P L U Y E N T A l ' C O H l / ^ N G D A N C A C H G I A I

Cau 6: M o t song co hpc co bien d p A, buoc song X. Toe d p dao d p n g eye dai

C a u 1: Song n g a n g t r u y e n dupe t r o n g cac m o i t r u o n g nao sau day :

ciia p h a n t u m o i t r u o n g bang 3 Ian toe dp t r u y e n song. T i m h ^ t h u e lien h?

A. Ran, l o n g .

giua A va X.

B. L o n g , k h i .

A.A=2TIA/3.

B . A = 27XA.

C. Ran va m a t tren m o i t r u o n g chat long.
Huang dan

gidi:

T
' A
o^
-1 AcoT
Ta CO A c j = 3— =>X =

27iA

T

3

Song ngang t r u y e n d u p e t r o n g m o i t r u o n g ran va tren m a t chat long.
5. i

=> C h p n C
A. RSn, long, k h i .

B. R^n, long.

C. K h i , chan k h o n g .

D. R^n, chan k h o n g .

A. 25 cm/s.
Chpn A

t r u y e n song.
B. Qua t r i n h t r u y e n song k e m theo s u t r u y e n nang l u p n g t u n g u o n t o i

D. V ^ n to'c t r u y e n song la v a n toe t r u y e n pha dao d p n g .

B. 50 cm/s.

C.

Huang dan

100 em/s.

D . 150 em/s.

gidi:

K h o a n g each giCra 7 g p n l o i lien tiep la 3 cm: 6A = 3 em => A = 0,5 c m
Toe d p t r u y e n song tren m a t nuoe bang v = A . f = 0,5.100 = 50 cm/s.

A. Q u a t r i n h t r u y e n song k e m theo s y v a n c h u y e n vat chat theo p h u o n g

C. Q u a t r i n h t r u y e n song la qua t r i n h t r u y e n pha dao d p n g .

^.
.
=> C h p n A

cac song t r o n Ian r p n g tren mat nuoe. Bie't k h o a n g each giira 7 g p n l o i lien

Cau 3: Ket l u a n nao sau day khong dung ve s u t r u y e n song co

n h i j n g cho t r o n g m o i t r u o n g ma song t r u y e n t a i .

gidi:

tiep la 3 c m . Toe d p t r u y e n song tren m a t nude bang

gidi:

Song dpe t r u y e n d u p e t r o n g cac m o i t r u o n g rin, long, k h i .

3

D . A = 3nA/2

X a u 7: N g u o n phat song S tren mat nuoe tao dao d p n g v o i tan so f = lOOHz gay

C a u 2: Song dpe d u p e t r u y e n t r o n g m o i t r u o n g nao d u o i day:

Huang dan

C . A = 37TA/4.

Huang dan

D. RMn, l o n g , k h i .

194

gidi:

=> C h p n B
Cau 8: M o t d i e m O tren m a t nuoe dao d p n g v o i tan so 20 H z , v g n toe t r u y e n
Song tren m a t n u o e thay d o i t u 0,8 m/s < v < Im/s. T r e n m a t n u o e h a i d i e m A
Va B each n h a u 10 c m tren p h u a n g t r u y e n song l u o n dao d p n g n g u p c pha
' A a u . Buoc song tren m a t n u d e la:

11.

A. 4 cm.

B. 16 c m .

C.

25 cm.

D. 5 cm.

.,•<• .,
195


1^1 inuttigminnuTiHBins

ii, rapi-Nguym

varu

Huang

nuangLac

dan

Huang

gidi:

D p l§ch p h a d a o d p n g c u a h a i p h a n t u tren p h u o n g t r u y e n s o n g e a c h nhai.
d la: A(p = ^ | ^ ( r a d ) . T h e o g i a thiet
2 Ted
= K + k.2.7t = :

^ ^,
2d.f
= l + 2k=> V =
l + 2k

M a 0,8 m/s < V < 1 m/s -> 0,8 m/s <
l + 2lc

^

A = 0,04 m = 4 c m . ^

B u o c s o n g la A = y ; v k h o n g d o i n e n k h i t a n g f l e n 3 I a n thi b u o c s o n g g i a m
3 Ian. => C h p n B.

2d.f

A(p = 71 + k.2n =>

m
l + 2k

< l m / s ^ l , 5 < k < 2 ^ k

g

= 2

g^ajH-

P h a t b i e u n a o s a u d a y v e d a i l u p n g d a c t r u n g c u a s o n g c o h p c la

khong diing?
A. C h u k y ciia song chinh bang c h u ky dao dpng cua cac p h a n t u dao dpng.
B. T a n so c i i a s o n g c h i n h b 3 n g tan so d a o d p n g c i i a c a c p h a n t u d a o d p n g .
C . T o e d p c u a s o n g c h i n h b a n g toc d p d a o d p n g c u a c a c p h a n t u d a o d p n g .

Chpn A

D. B u o c s o n g l a q u a n g d u o n g s o n g t r u y e n d i d u p e t r o n g m p t c h u k y .

Huang dan gidi:

C a u 9; M p t s o n g c o h p c I a n t r u y e n t r o n g m p t m o i t r u o n g v a t c h a t tai m p t diei
each n g u o n

x (m)

co p h u o n g

trinh song:

u = 4cos
3

6

c m . V a n toc

B. I m / s .

T r o n g c a c d a i l u p n g d a c t r u n g c u a s o n g thi: C h u k y c i i a s o n g c h i n h b a n ?
chu k y d a o d p n g c u a c a c p h a n t u d a o d p n g ; T a n so c i i a s o n g c h i n h b a n g tan so
dao d p n g c u a c a c p h a n t u d a o d p n g ; C o n toe d p t r u y e n s o n g l a toc d p t r u y e n

t r u y e n s o n g t r o n g m o i t r u o n g co g i a tri:
A . 2 m/s.

dan

C . 0,4m/s.

D . M p t g i a tri khac.

pha d a o d p n g v a k h o n g b a n g toc dp d a o d p n g c u a c a c p h a n t u d a o d p n g . .
.a

Chpn C.

Huong dan gidi:

C a u 14: M p t s o n g c o I a n t r u y e n tren m p t d u o n g t h a n g t u d i e m O d e n d i e m M
Ta

C O to = — ;

3

T = 6s;A = 2 , 4 m ^ v = ^

6

= 0 , 4 m/s

trong q u a t r i n h s o n g t r u y e n . N e u p h u o n g t r i n h d a o d p n g c u a p h a n tit v a t c

•=> C h p n C
C a u 10: K h i m p t s o n g c o h p c t r u y e n t u k h o n g k h i v a o n u o c thi d a i l u p n g nao

tai d i e m M co d a n g U j ^ ( t ) = a sin27rft thi p h u o n g t r i n h d a o d p n g c u a p h a n ti'*V9t c h a t tai O l a :

sau day khong thay doi:
A . V a n toc.

each O m p t d o a n d. Biet tan so f, b u o c s o n g X v a b i e n d p a c u a s o n g k h o n g doi

B. T a n so'.

C . B u o c song.

D. N a n g lupng.

A. U Q ( t ) = a s i n 7 i

B. U Q ( t ) = a s i n 7 t ft +

C. Uo(t) = asin27t f t . ^

D. U Q ( t ) = asin27t

Huang dan gidi:
V a n toc t r u y e n s o n g t r o n g c a c m o i t r u o n g k h a c n h a u l a k h a c n h a u ; d o vay
k h i m p t s o n g c o t r u y e n t u m o i t r u o n g n a y s a n g m o i t r u o n g k h a c thi v a n toc
thay d o i , b u o c s o n g v a n a n g l u p n g c u n g t h a y d o i c h i co t a n so l a k h o n g thav
n o i . => C h p n B.
C a u 11: M p t s o n g c o co tan s o f, t r u y e n t r o n g m o i t r u o n g v a t cha't d a n h o i , voi

B.X = 2vf.
V

D.A=--.
f

Huang dan gidi:
T a CO buoc song A = j

X

Huong dan gidi:
P h u o n g t r i n h d a o d p n g c u a p h a n t u v a t c h a t tai O s o m p h a h o n tai M

v a n toe l a v thi b u o c s o n g la
A . ? . = vf.

X

la

t= ^
P h u o n g t r i n h d a o d p n g c i i a p h a n t i i v a t c h a t tai O l a
U Q (t) = asin27tf

t+

- = a s i n 271
V

= a s i n 271 ft +
V

X

=> C h p n C .

=> C h p n D .

£|u_15: M p t n g u o n p h a t s o n g d a o d p n g v o i p h u o n g t r i n h u = a s i n 2 0 7 t t ( c m )
C a u 12: M p t s o n g c o h p c I a n t r u y e n trong m o i t r u o n g d a n h o i v o i v a n toc
k h o n g d o i , k h i t a n g tan so l e n 3 I a n thi b u o c s o n g
A. T a n g 3 Ian.

B.Giam31an.

C . T a n g 9 Ian

^ o i t tinh b a n g g i a y . T r o n g k h o a n g t h o i g i a n 2s s o n g n a y t r u y e n d i d u p e q<. ~ 7

^irong b^ng bao n h i e u Ian b u o c song?

D.Giam91an.

A . 10.

B. 40.

C . 30.

D.20.

,

. A-

19/


Huang dan

gidi:

J a u 20: K h o a n g each g i i i a hai n g o n song lien tie'p tren m a t ho la 3 m . T r o n g
jyipt p h i i t song d a p vao b o 7 Ian. V a n {oc t r u y e n song la

to

Tan so song la f = — = l O H z ; V a n toe song la v = A.f = lOX
271

Q u a n g d i r o n g la S = v.t = 10X.2 = 20X

Chpn D .

A . 0,6 m/s.

B. 0,2 m/s.

C. 0,5 m/s.

Huang dan

ci JW/fv

C a u 16: M p t n g u o i q u a n sat m p t chiec la tren m a t n u o c thay n o n h o cao 8 JaY,

C h u k i eiia song la T = — = 10s;

trong 14 giay. K h o a n g caeh g i u a hai n g p n song lien tie'p la 2 m . V a n to'c truyen

D . 0,3 m/s. '

gidi:

V a n to'c t r u y e n song v = — = 0 , 3 m / s

6

song la

T

Chpn D .

A. Im/s.

B. 2m/s.

-»

C. 3m/s.

Huang dan

D . 4m/s.

Cau 21: Song tai m p t d i e m M dao d p n g v o i chu k i T = 2s, v a n to'c t r u y e n song la

gidi:

V = 0,5 m/s. K h o a n g each giiJa hai d i e m ngupe pha nhau gan nha't la

Chiec la n h o cao 8 Ian, tuc la thuc h i ^ n d u p e 7 chu k i t r o n g 14 giay

A . 1,5m.

B. I m .

=> C h u k i dao d p n g eiia chiec la la T = 2s;

Huang dan

V a n to'c t r u y e n song la v = A/T = 1 m/s . => C h p n A .
chu k i 0,5s. Buoc song cua no la
B. 3 m .

C. I m .
Huang dan

D . 0,5 m .

gidi:

Buoc song A, = v T = l m

C a u 17: M p t song co hpe t r u y e n t r o n g m o i t r u a n g d a n h o i v o i v a n toe 3m/s va
A . 1,5m.

C. 2 m .

D . 2m.

Khoang each giij-a hai d i e m dao dpng ngupe pha gan nhau nha't

^ = 0,5m

=> C h p n D

gidi:

Cau 22: M p t spi day dan h o i n a m ngang co m p t d a u co' d i n h , m p t d a u dao

Ta C O b u o c song la A. = v T = 1,5m

dpng v o i bien d p a = 10cm va chu k i T = I s . C h p n go'e t h o i gian t = 0 k h i vat d i

=> C h p n A .

qua v j t r i co l i d p bang 0 theo chieu d u o n g . P h u o n g t r i n h dao d p n g eiia song la

C a u 18: P h u o n g t r i n h n g u o n song tai O la

UQ =

4COS

lOOTtt-2

(cm). Bie't van

to'c t r u y e n song la v = 5m/s. P h u o n g t r i n h song tai M each O m p t d o ^ n 5cm la
A.

Uj^

= 4 cos l O O T i t - 2

C. u ^ = 4 cos 1007tt + 2
Huang dan

,

Huang dan

D . u ^ =4cosl007it

Khi t = 0 :

U(3

= 0;

Vj)

> 0 . Ta

CO

gidi:

(p = - |

P h u o n g t r i n h dao d p n g eiia song la u = 1 0 c o s ( 2 7 r t - T c / 2 ) ( e m ) . C h p n A .

gidi:

500

= 4 cos lOOTlt-

3n

= 4 cos 1007tt + 2)

cm

C a u 19: M p t song c6 p h u o n g t r i n h u = 3eos(57tt + 0,047:x) ( e m ) , t r o n g do >
d u p e t i n h b a n g c m , t d u p e t i n h bang giay. Buoc song eiia song do la
B. 40em.

C. 60cm.

Huang dan
2nx
T a c o 0,04Ttx = -

D . u = 10cos(27it + 7i) ( m ) .

T a n s o a) = - ^ - 2 7 i ( r a d / s )

=> C h p n C.

A . 70cm.

C. u = 10cos(27it + n/2) (em).

£ a u 23: T r e n mat m p t chat l o n g tai O co m p t n g u o n song co dao d p n g v o i tan

1
200

B. u = 10cos(27:t - n/2) ( m ) .

B. U j ^ = 4 c o s ( l 0 0 r t t - 7 i )

P h u o n g t r i n h song tai M la
u ^ = 4cosl007i t - -

A . u = 10eos(27it - 7i/2) (cm).

X = 50cm

gidi:

Chpn D .

D . 50cm.

so f = 30Hz, van toe t r u y e n song la m p t gia t r j t r o n g k h o a n g l , 6 m / s < v <
2,9m/s. Bie't tai d i e m M each O m p t k h o a n g lOem, song tai d o l u o n dao d p n g
nguoc pha v o i dao d p n g tai O . Gia t r i van toe do la
A . 2m/s.

B. 4m/s.

C. I m / s .

,
D . l,2m/s.

Huong dan gidi:
-

(i

Dao d p n g tai M l u o n n g u p e pha v o i dao d p n g tai O nen
A(p = 7t + 2k7r =

27td

=> V =

2df

_

6

fm^

2k + l ~ 2k + l l s


Theo bai ra 1,6 m/s < v < 2,9 m/s => 1,6 < — ^ — < 2,9
2k+ 1

Huang dan

=> k = 1

gidi:

pao dpng tai M luon ciing pha voi dao dpng tai N
27td

nen ^ 9 - 2 k T : = —

V a y v a n toe la v = 2m/s. =i> C h o n A .

2df

=^ v = —

9 ^m^

=

-

•

'37rt + ^ ' cm

Cau 24; M p t n g u o n song co dao d o n g theo p h u o n g t r i n h u = A cos V

2;

t r o n g do t d o bang giay. K h o a n g each giua hai d i e m gan nhat tren p h u o n ^
t r u y e n song ma pha dao d o n g lech pha nhau Zn/l

la 0,75m. Buoe song va van

toe t r u y e n song Ian l u g t la
A. lm;l,5m/s.

X

Xheo bai ra 0,7 m/s < v < 0,8 m/s => 0,7 < — < 0,8 => k = 6.
Van toe la v = 0,75m/s = 75cm/s => Chpn A .
Cao 27: Song co t r u y e n t r o n g m o t m o i t r u o n g dpc theo true O x v o l p h u o n g

B. 2,5m;lm/s.
Huang dan

T a c o A(p =

C. 0,75m; l,5m/s.

D . 1,5m; 5m/s.

gidi:

' ^ ^ ^ ^ = cos(20t - 4x) (cm) (x t i n h bang met, t t i n h bang giay). V a n toe t r u y e n
song nay t r o n g m o i t r u o n g tren bSng
A. 5 m/s.

= — =>X = — = l m
2
3

4x =

Taco:

Cau 25: Tai d i e m S tren m a t n u o c yen t i n h co n g u o n dao d o n g d i e u hoa theo
p h u o n g thang d u n g v o i tan so f. K h i do tren mat n u o c h i n h thanh he song tron
d o n g t a m S. Tai hai d i e m M , N nam each nhau 5em tren d u o n g thang d i qua S
l u o n dao d p n g nguge pha v o i nhau. Biet toe d p t r u y e n song tren m a t n u o c la
80cm/s va tan so ciia n g u o n dao d p n g thay d o i t r o n g k h o a n g t u 48Hz den
64Hz. Tan so dao d o n g ciia n g u o n la
B. 48Hz.

'»

Huong dan

C. 54Hz.

D.56Hz.

gidi:

D p lech pha dao d p n g ciia hai phan t u tren p h u o n g t r u y e n song each nhau
,

,

a la: A(p =

27rd

=

27td

,

.f

Cau 28: Song co co tan so 80 H z Ian t r u y e n t r o n g m o t m o i t r u o n g v o i van toe
4m/s. Dao d p n g ciia eac phan t u vat chat tai hai d i e m tren m o t p h u o n g t r u y e n
song each n g u o n song n h u n g doan Ian l u p t 31cm va 33,5cm, lech pha nhau goe
B. n rad

A. — rad
2

2,5 < k < 3,5;

Cau 26: Tai d i e m S tren m a t nuoc yen t i n h co n g u o n dao d p n g dieu hoa theo
p h u o n g thang d u n g v o i tan so 50Hz. K h i do tren m a t n u o c h i n h thanh h$ song
t r o n d o n g tam S. Tai hai d i e m M , N nSm each nhau 9em tren d u o n g th3ng d i
qua S l u o n dao d p n g ciing pha v o i nhau. Biet rang, toe do t r u y e n song thay doi
t r o n g khoang t u 70cm/s den 80em/s. Toe d p t r u y e n song tren m a t n u o c la
B.80cm/s.

C. 70em/s.

C. 2n rad

Huang dan
V
400
X=—=
Ta CO •
f
80
,

=

^
5(cm
^ ^

27td

27t.2,5

X

5

=

D. - r a d

gidi:

Chpn B

/

,s

nyradj

M o t song t r u y e n theo true Ox v o i p h u o n g t r i n h u = acos(47:t - 0,02nx)

A . 100 cm/s.
48 H z < 8(2k + 1) < 64 H z

5m / s

=> Chpn A.

5

V i k n g u y e n nen k = 3 -> f = 56 H z . ^ Chpn D

A.75cm/s.

T

va X t i n h bang cm, t t i n h bSng giay). Toe d p t r u y e n ciia song nay la

2.5

< • M a 48 H z < f < 64 H z

D.4 m/s.

gidi:

V = — =

(0 = 20

£la^:

^ (2k + l ) . v
(2k + l).80
,
,
f = ^ — / - = ^ - — / - = 8(2k.l)(Hz)
2d

C.40 cm/s

X = 0,5n

271X

Acp = —;— = — - —

Theo gia thiet A(p = TT + 2k7i nen ^ ^ . f = 71 + 2k7i
v
•"

B.50cm/s.
Huang dan

V a n toe t r u y e n song v = Xf = X — = l , 5 m / s =i> ChQn A .
2n

A . 64Hz.

J-

D.72em/s.

Ta co: <

B. 150 cm/s.
Huang dan

2n\
0,027tx = X
CO = 47t

?. = 100(cm)
T = 0,5(s)

C. 200 em/s.

D.50 cm/s.

gidi:
. v = Y = 200cm/s

^ChpnC.

^Sa_30: M o t song co co chu k i 2 s t r u y e n v o i toe d p 1 m/s. K h o a n g each giua
d i e m gan n h a u nhat tren m p t p h u o n g t r u y e n m a tai do eac phan t u m o i
^'^^Ong dao d p n g ngupe pha nhau la
A . 0,5m.

B. 1,0m.

C.2,0m.

D . 2,5 m .


Huang dan
X = v.T
Ta c6:

, = 2(m)

^

d=l(m)

2

gidi:

Huang dan

gidi:

H a i d i e m tren c i m g m o t p h u o n g truyen song, each nhau m o t khoang bang
C h o n B.

]piidc song t h i dao d o n g c i i n g pha.-> Chon A
0 j j _ 3 5 : M o t song h i n h sin truyen theo p h u o n g Ox tii n g u o n O v o i tan so 20Hz,
toe do truyen song n a m trong khoang tir 0,7 m/s den 1 m/s. Goi A va B la hai

C a u 31: Buoc song la khoang each giua hai d i e m
A . tren c u n g m o t p h u o n g t r u y e n song ma dao d o n g tai hai d i e m d o ng^^^^

^il'm n a m tren Ox, 6 c i i n g m o t phia so v o i O va each nhau 10 em. H a i phan tir
jfldi t r u o n g tai A va B l u o n dao d o n g nguge pha v o i nhau. Toe d o t r u y e n song la

pha.
B. gan nhau nhat tren cung m o t p h u o n g t r u y e n song m a dao d o n g tai hg,

A.lOOem/s

B. 80 em/s

C. 85 em/s

Huong dan

d i e m d o Cling pha.
C. gan nhau nhat ma dao d o n g tai hai d i e m do cung pha.
Huang dan

gidi:

27:d

oi

Acp = 2k7r + 71 =

D . tren ciing m g t p h u o n g truyen song ma dao dong tai hai d i e m d o ciing pha,

D . 90 cm/s

X

=>0,7<2k + l < l = > k = l=> v = 0 , 8 m / s

o,7
gidi:

K h o a n g each giira hai d i e m tren cung m o t p h u o n g t r u y e n song ma dao
d o n g tai hai d i e m d o n g u o c pha la nua b u o c song -> D a p an A sai

'

-»ChgnB.
Cau 36: H a i d i e m M , N ciing n a m tren m g t h u o n g t r u y e n song va each nhau

Buoc song la k h o a n g each giua hai d i e m gan nhau nhat tren cung rrmt

mot phan ba b u o c song. Bien d o song k h o n g d o i t r o n g qua t r i n h t r u y e n . Tai

p h u o n g t r u y e n song m a dao d o n g tai hai d i e m d o c u n g pha. D a n an B dung;

mot thoi d i e m , k h i l i d o dao d o n g ciia phan tir tai M la 3 cm t h i l i do dao d o n g

C va D sai. => C h o n B.

cua phan tir tai N la - 3 cm. Bien do song bang

C a u 32: M o t nguon phat song co dao dong theo p h u o n g trinh

u=4oo6

4

Biet dao d o n g tai hai d i e m gan nhau nhat tren ciing m o t p h u o n g t r u y e n song
each nhau 0,5 m c6 d o lech pha la n/3. Toe d o t r u y e n cua song d o la :
B. 2,0 m/s.

A . 1,0 m/s

C. 1,5 m/s.

Huang dan

D . 6,0 m/s.

3
(0

X
= 471

B. 3 em.
Huang dan

/,=>v
T = 0,5(s)

= —= 6m/s=>
T

a. cos cos (CO t ) = 3
a. coscos c o t - -

271"!

_=>a = 2 N / 3 ( c m ) ^

=

-3

X
v = - = X{ = 15m/s
T

A l l .

B. 8.

>

->ChonB.

C. lech pha n/2

Ta

CO

f

{

—
X

p h u o n g t r u y e n song, each nhau m o t khoang bang b u o c song co dao d g n g
B. N g u g e pha.

C. 5.

D.9.

gidi:

D . 25 m/s

C a u 34: M o t song co Ian t r u y e n t r o n g m o t m o i t r u o n g . H a i d i e m tren cung

A . C i i n g pha.

S2 each nhau 8,2cm,

song tren mat nude la 30cm/s, coi bien d o song k h o n g d o i k h i t r u y e n d i . So

Huang dan
C. 30 m/s

Huang dan gidi:
f = 120Hz

ChgnC.

<^'em dao d o n g v o i bien d o cue dai tren doan 8582 la

n a m 0,5 m . Toe d o t r u y e n song la

Ta co: <

gidi:

fhang d u n g co tan so 15Hz va l u o n dao d o n g d o n g pha. Biet van toe t r u y e n

120Hz, tao ra song on d j n h tren mat chat long. Xet 5 g g n 161 lien tiep tren mP'

B. 15 m/s

D . 3N/2 cm.

"guoi ta dat hai n g u o n song co ket h g p , dao d o n g d i e u hoa theo p h u o n g

ChonD.

p h u o n g t r u y e n song, 6 ve m o t phi'a so v o i n g u o n , g g n t h i i nhat each ggn thif

4?. = 0,5

cm.

£|u 37:Tren mat n u o c n a m ngang, tai hai d i e m S^,

C a u 33; Tai m o t d i e m tren mat chat long co m p t n g u o n dao d o n g v o i tan so

A . 12 m/s

C. 2 ^

gidi:

27id
Ta c6:

A. 6 em.

D . I?ch pha n/4

15

^

^

= > - ^ < k < . ^ = ^ k = ±4, + 3 , ± 2 , ± l , 0


Chgn D
^^JL38: De khao sat giao thoa song ca, n g u d i ta bo' t r i tren mSt nude n a m
hai n g u o n ket h g p Si va 82. H a i n g u o n nay dao d g n g dieu hoa theo


^

^i^y

p h u o n g thSng d u n g , c u n g pha. X e m bien d p s o n g k h o n g thay d o i t r o n g
t r i n h t r u y e n song. Cac d i e m thupc mat nuoc va n a m tren d u o n g t r u n g t r y c ci;i^
doan S1S2 se

,

A . dao d p n g v o i bien d p cue d a i .
i

B. dao d p n g v o i bien d p cue tieu.

t

C. k h o n g dao d p n g .

,Y>if,. ,^

it;

^ ^ D . mpt so le Ian b u o c song.
,

bien dp cue dai.

mat n u Q c va n a m tren d u o n g t r u n g t r y c ciia dogn S1S2 c6 d2 - d i = 0 se dao
d p n g v o i bien d p cue d a i . -> C h p n A
Cau 39: Tai hai d i e m M va N t r o n g m p t m o i t r u o n g t r u y e n song c6 hai nguCin
song ket h p p cung p h u o n g va cung pha dao d p n g . Biet bien d p , v a n toe cua
song k h o n g d o i t r o n g qua t r i n h t r u y e n , tan so' ciia song bang 40 H z va c6 sir
giao thoa song t r o n g doan M N . T r o n g dpan M N , hai d i e m dao d p n g c6 bien do
cue d a i gan n h a u nhat each nhau 1,5 cm. V a n toe t r u y e n song t r o n g m o i truong
nay bang
•

B. 1,2 m/s.
'

Huang dan

C. 0,3 m/s.

gidi:

txx hai n g u o n den d i e m do bang mpt so' nguyen Ian buoc song se dao dpng v o i

gidi:

H a i n g u o n cung pha, cac c u e dai giao thoa c6 d2 - d i = kA -> Cac d i e m thuoc

A . 2,4 m/s.

Huang dan

^PlHai n g u o n dao dpng c i m g pha nen n h u n g d i e m eo hi?u d u o n g d i ciia song

D . dao d p n g v o l bien d p bang nua bien dp cue d a i .
Huong dan

'p. mpt so n g u y e n Ian buoc song.
C. mpt so' n g u y e n Ian n u a buoc song.

• fj.;, w v; , > "
,j

"'^^ b u o c song,

j A-

D.0,6 m/s.

nhau 20cm. H a i n g u o n nay dao dpng theo p h u o n g thang d u n g eo p h u o n g
trinh Ian lupt la U j = 5eos407it ( m m ) va U 2 = 5eos(407it + n) ( m m ) . Toe dp
truyen song tren m a t chat l o n g la 80 cm/s. So d i e m dao dpng v o i bien dp cue
dgi tren doan t h ^ n g S1S2 la:

A. 11.

B.9.

C.IO.
Huang dan

D.8.

'

gidi:

j^jjIlHai n g u o n dao dpng ngupc pha nen n h u n g d i e m dao dpng v o i b i e n dp eye
dgi tren doan thang S1S2:
80
.
V
X = — = — = 4(em)

\

J

J

X

2

=^ -5,5 < k < 4,5 r:^ = - 5 , +4, ±3, +2, ±1, 0 => C h p n C


Ta CO • 2 " ^'^('^'") ^ V = 3.40 = 120(cm) - 1,2m => C h p n B.
Cau 40: Tai hai d i e m A va B t r o n g m p t m o i t r u o n g t r u y e n song c6 hai

O be m a t m p t chat l o n g c6 hai n g u o n phat song ket h p p Si v a S2 each

0aJ2:

gidi:

v = Xf

C h p n B.

X

2

Cau 43: 6 mat thoang ciia mpt chat long eo hai n g u o n ket h p p A va B dao dpng
dieu hoa c i i n g pha v o i nhau va theo p h u o n g thang d u n g . Biet toe dp t r u y e n

n g u o n song ket h p p , dao d p n g c i i n g p h u o n g v o i p h u o n g t r i n h Ian luot la

song k h o n g d o i t r o n g qua t r i n h Ian truyen, buoc song do m o i n g u o n tren phat

u ^ = acoscot va Ug = 2acos((ot + n). Biet van toe va b i e n d p song d o m o i nguon

rabang 12 cm. K h o a n g each ngan nhat giira hai d i e m dao dpng v o i bien dp eye

tao ra k h o n g d o i t r o n g qua t r i n h song t r u y e n . T r o n g k h o a n g g i u a A va B c6

liai nam tren doan th^ng A B la

giao thoa song d o h a i n g u o n tren gay ra. Phan t u vat chat tai t r u n g d i e m cu'a

A. 9 em.

doan A B dao d p n g v o i bien d p bang
A.O

B.a/2

C.6cm.

Huang dan
C.a

Huang dan

B. 12em.

D.2a

gidi:

gidi:

Khoang each ngan nhat g i i i a hai d i e m dao d p n g v o i bien dp eye dai n a m
^4^44: D i e u kien de h a i song co k h i gap nhau, giao thoa dupe v o i nhau la hai

Chpn C

Cau 41: O m a t nuoc eo hai n g u o n song dao d p n g theo p h u o n g v u o n g goc
mat nuoc, eo cung p h u o n g t r i n h u = Acoscot. T r o n g m i e n gap nhau ciia IT"'
song, nhOng d i e m ma 6 d o cac p h a n tir nuoc dao d p n g v o i bien d p eye dai
CO hi§u d u o n g d i cua song tix hai n g u o n den do bang

M

'fen doan th^ng A B la d = A/2 = 12/2 = 6 (cm). ^ Chpn C

H a i n g u o n dao d p n g ngupc pha, nen tai t r u n g d i e m cua doan A B '-'^
d2 - d i = 0 se dao d p n g v o i bien d p cue tieu A = I ai - a21 = a.

D . 3 cm.

phai xuat phat t u hai n g u o n dao dpng

C

A. cung bien dp va eo hi^u so pha k h o n g d o i theo t h o i g i a n
eiing tan so, cung p h u o n g

. CO ciing pha ban dau va ciing bien dp

. c i m g tan so', c i m g p h u o n g va co hieu so'pha k h o n g doi theo t h o i gian

Huang dan

Huang dan

gidi:

D i e u ki§n de hai song co k h i gap nhau, giao thoa d u g c v o i n h a u la hai song
phai xuat phat t u hai n g u o n dao d o n g ciing tan so^ c i i n g p h u o n g va c6 h i ^ u SQ
pha k h o n g d o i theo t h a i gian. ^ C h o n D
Cau 45: 6 m a t chat l o n g c6 hai n g u o n song A , B each n h a u 20cm, dao dong
theo p h u o n g t h i n g d u n g v o i p h u o n g t r i n h la u ^ = Ug = 2cos507tt (t t i n h bang
s). Toe do t r u y e n song tren mat chat long la l,5m/s. Tren doan thang A B , so
d i e m CO bien d p dao d o n g cue dai va so d i e m d u n g yen Ian l u p t la
A. 9 v a 8

B. 7 v a 8

C. 7 va 6

Huang dan

D . 9 va 10

gidi:

X

Bien d p dao d p n g tai M : A ^ = 2a c o s 7 i . ^ \ ~ ^ ^ = 2.2 c o s T t . - = 2V2

(cm)

4

X

^ Chpn B.

0ii48: H a i

.

n g u o n ket h p p Si va S2 each nhau 25 c m dao d p n g v o i tan so 25 H z ,

Cling pha tao hai song giao thoa v o i nhau tren mat nuoe . V a n toe t r u y e n song
1^ 1,5 m/s. GiiJa S1S2 co bao nhieu gpn loi?
A. 7
B.9
C.5

D.4

gidi:

Ta CO

^ =^7 =^50
^

.

V
150
^/
^
X= - =
= 6(cm)

—

puoc song X = - = — = 4 { c m )

Huang dan

Ta CO

gidi:

f
25

^

^

X

11

k = +3,±2,±l,0

-3,8 < m

m = -3,±2,±l,0

<2,8

^ Ch(?n C

11


1 2

-3,3
X

,
.
= 6(em)

I
,
— X

- 4 , 2 < k < 4 , 2 = > k = ±4,±3,+2,±l,0

X

=> Chpn B
Caai9: Tai hai d i e m A va B tren mat nuoe dao d p n g e i m g tan so 16HZ, cung

2

Cau 46: Tai mat thoang ciia m p t chat long c6 hai n g u o n song Si va S2 dao dong

pha, Cling bien d p . D i e m M tren mat nuoe dao d p n g v o i bien d p cue dai v o i

theo p h u o n g thSng d u n g v o i cung p h u o n g t r i n h u = aeos407it (a k h o n g doi, t

MA = 30 cm, M B = 25,5 em, giifa M va d u o n g t r u n g true cua A B eon co hai day

t i n h hang s). Toe do t r u y e n song tren mat chat long bang 80 cm/s. K h o a n g each

ci^c dai khae. Van toe truyen song tren mat nuoe la :

ngan nha't g i u a hai p h a n t u chat long tren doan t h i n g S1S2 dao d p n g v o i bien

A . 36 cm/s.

C. 20,6 cm/s.

Huang dan

d p cue dai la
A . 4 em.

B. 6 cm.
Huong dan

.

B. 24 cm/s.

V

80

./

C. 2 cm.

D . 1 cm.

gidi:

X= ^ =
f

d, -d2

^

= 6(em)_I

25 — f,i

=k?.

gidi:

k =—3n
30'-25,5-3X

X = 1,5(em) => V = 24cm / s

=> Chpn B

^ Chpn C

Ta CO

Ta CO <

150

-^U-50: Tren mat nuoe, hai n g u o n ket h p p each nhau 40cm dao d p n g cung

d = ^ = 2(cm)

P 9/ ^ = 6cm. H a i d i e m C D nam tren mat nuoe ma A B C D la m p t h i n h chix nhat

Cau 47: Tai m a t chat long co hai n g u o n phat song ket h p p Si va S2 dao dpng

^ * 30em. So d i e m cue dai va d u n g yen tren C D la

theo p h u o n g v u o n g goe v o i mat chat long co cung p h u o n g t r i n h u = 2cos40n'

B.7;6.

(trong do u t i n h bang cm, t t i n h bang s). Toe d p t r u y e n song tren m a t chat long
la 80em/s. G p i M la d i e m tren mat chat long each Si,S2 Ian l u p t la 12cm va 9cm

C. 13;12.
Huang dan

gidi:

'AD-BD
t u chat l o n g tai M dao d p n g v a i bien dp la

AD-BD<(m+0,5)>.
B. 2>/2cm

C. 4em

D . 2em

D . 11;10.

Ta CO

Coi bien d p cua song t r u y e n t u hai n g u o n tren den d i e m M la k h o n g d o i . Pha"^
A . 42 em

D . 28,8 cm/s.

^ChpnB

f-3,3
k = -3,...,3
m = -2,...,3
r/>o


Cau 51: C h o hai n g u o n ket h g p Si,S2. M , N n a m tren d u a n g t h a n g Si,S2
h l n h ve. Tai N la v a n cue d a i . Tai M la van

Huang dan
IK,

QiVi 55: Q u a n sat hien t u g n g giao thoa tren mat n u a c d u g e t^o t h a n h do hai

'i' e

C. T r u n g gian.

S2

'

D . K h o n g d i i d i e u kien xae d j n h .

N

pguon ket h g p A v a B c u n g pha dao d g n g v o i tan so 15Hz, n h a n tha'y

ng c6

bien d g cue d a i bac nhat ke t u d u o n g t r u n g true ciia A B , la tai n h u n g d i e m eo

'

Huang dan
Ta CO INS^ - NS2

H a i song U A va U B n g u g c pha nhau nen tai t r u n g diein song c6 bien d p bang 0
^ Chgn D.

A . Cue d ^ i .
B. Cue t i e u .

gidi:

j^i|u k h o a n g each d e n A va B bang 3em. V a n toe t r u y e n song t r e n m a t n u o e la

gidi:

A . 90cm/s.

B. 45cm/s.

M S , -MS2

C. 30cm/s.

D . 60cm/s.
j ' , . ^

Huang dan

Tai N la v a n cue d a i t h i tai M eung la van cue dai. -> C h g n A .
Cau 52: Tren mat n u d e eo hai n g u o n song Si va S2 dao d p n g cung pha ciing taV,

D o song CO bien d g cue d a i bac nhat nen k = 1
d2 - d , = k?i = 3 =>

^'^>'ju,i

= 3cm; v = ^if = 45cm / s => C h g n B.

^

so'va c u n g p h u o n g con bien d g song khae nhau. Bien d o song tai n h i r n g dietn

g | u 5 6 : H a i n g u o n phat song ket h g p A va B c i i n g tan so, c i i n g bien d g va c i i n g

ma h i ^ u k h o a n g each den hai n g u o n bang m o t s o l e Ian n u a b u o c song la

pha. Coi bien d g song k h o n g d o i . D i e m M , A , B, N theo t h u t u thang hang.

A . bang 0.

C. cue tieu.

Bien d g dao d g n g t o n g h g p tai M c6 gia trj 6 m m t h i bien d g dao d g n g t ^ ^ h ^

B. cue d a i .

D . k h o n g xae d i n h d u g c .

t^i N CO gia trj la

Huong dan

gidi:

C h g n d a p an C, v i hai n g u o n song k h o n g cung bien d o nen tai n o i dao dong

A . chua d u du: k i ^ n de t i n h .

B. 3 m m .

C. sVScm.

D. 6mm.

cue tieu song t o n g h g p v a n con dao dgng, v o i bien d o b i n g h i ^ u bien do cua
hai song thanh p h a n ma k h o n g phai la bang 0, A = A , - A2 ^ 0 .-> C h g n C.
Cau 53: H a i n g u o n song nuoe A va B gio'ng nhau each nhau 12cm dang dao
d g n g dieu hoa v u o n g goe v o i mat nuoe. Buoc song la 1,6cm. M la m o t diem
tren mat n u o e each deu 2 n g u o n 1 khoang 10cm. O la t r u n g d i e m A B . So dium
dao d g n g n g u g c pha v o i n g u o n tren doan O M la
A . 3.

B.5.

C. 4.
Huang dan

gidi:

n g u o n k h i c h i i n g each hai n g u o n mot khoang bang m o t s o l e Ian nua buoc song
AO < X < A M
keZ

71

u ^ = aeos cot + —
2

va Ug = acos c o t - ^
2

A

B

N

AOR- t i n s o r i X

6mm. ^ C h g n D .

^o'"^

^'^ §"02

A . Chieu d a i day bang m o t phan t u bu6c song.

'

B. Buoc song gap d o i chieu d a i day.

-

'

'nl

C. Chieu d a i day bang bgi so n g u y e n Ian n u a b u o c s o n ^ ' '
'
/
D. C h i e u d a i day bang m g t so le buoc song.
,x

giat:

"

'A .h

• r> rg? n*^' j
,A

.i\mdl

C. 3a/2.

D . 0.

. ,13 G ^

1

^ •

- ij

£au 58: Song d u n g , hieu so pha cua p h a n t u d o i x u n g qua n i i t la gr, 0:.- — |
A . 0.

B. 71.

C.27t
Huang dSngiAi:'-'

d g n g v o i bien d o
B. a.

^

bang m g t so n g u y e n Ian nua buoc song. -> C h g n C.

z:i> C h g n D .

T r o n g k h o a n g S1S2 eo giao thoa song xay ra, phan tir tai t r u n g d i e m S1S21^'^'
A . 2a.

in iRb o G .G

'

Song d i m g xay ra tren day dan h o i eo hai d a u eo d i n h k h i chieu d a i day

Cau 54: Cho hai n g u o n ket h g p Si, S2 c6 p h u o n g t r i n h Ian l u g t la:
.

Bien d g dao d g n g t o n g h g p tai N c6 gia

j.iRorl>J .3

i s , ^ > ^ ^

tri b3ng bien d g t o n g h g p tai M va hang

,

^ k = 4;5

u

-

Huang dan

6<(2k + l ) ^ < 1 0
3,25
Ta CO M A - M B = N A - N B = A B

-7T rr

Cau 57: Song d u n g xay ra tren day d a n hoi c6 hai dau co d i n h k h i ' "
D.2.

V i hai n g u o n gio'ng nhau nen nhOng d i e m tren O M dao d g n g ngugc pha vm
Ta CO

Huang dan gidi:

.^^^

D.jr/2.
f - "

V -

«' ^OM : £ d n i 2

Cac phan t u d o i x u n g qua n i i t t h i n g u g c pha nhau nen hi^u so p h a A y ^^a
mh u^i^

ChgnB.

„

^

.A
_