7. sound wave

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Rise Academy, 607/608-A, Talwandi , Kota (Raj.) -324005. 134

EXERCISE # 1

1.4 IysV }kjk nwjh r; djus es a yxk le;

10 102 =2

1gt2 =

2

19.8t2 t =

7

1sec.

f =yxk le;

kdh la[;dEiuks=

7/1

8= 56 Hz3

2.2* V T V increases c

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laiks"kh O;frdj.k ds fy, iFkkUrj (S)S = n (n = 1, 2, 3, 4....... )

;gk S = 2 mvr% laiks"kh O;frdj.k ds fy, fodYi A vkSj B fodYi lgh gS vr% blh rjg fouk'kh O;frdj.k lehdj.k ds fy,

S = (2n + 1)2

vr% D fodYi lgh gSA Ans A, B, D,

4.4 First maxima after O will appear when path difference S = so AP BP =

22 14.2 2.4 = = 0.2

sound velocity = n = 1800 0.2 = 360 m/sO ds ckn izFke mfPp"B izkIr gksxk tc iFkkUrj S = gSAvr% AP BP =

22 14.2 2.4 =

= 0.2

/ofu osx = n = 1800 0.2 = 360 m/s4.8 path difference iFkkUrj = r 2 r

S = r ( 2) n = Sfor constructive interference

lEiks"kh O;frdj.k ds fy;s

n = r ( 2) =n

)2(r n =

V

= )2(r

Vn

6.2 n1: n

2: n

3= 1 : 2 : 3

1

V

: 2

V

: 3

V

= 1 : 2 : 3

1

: 2

: 3

= 1 :21

:31

6.62

1= + .6d,

1=

1

V

4

2= + .3d,

2=

2

V

1

2

= )d3.(4

)d6.(2

= )d3.(2

)d6.(

6.15 Equation of stationary wave from closed end = a sink x cos tcan fljs ls vizxkeh rjax dk lehdj.k = a sink x cos t

Now at x =7

4

7=

Amplitude vk;ke = a sin kx = a sin7

2

= a sin74

72

= a sin2

= a

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7.2* 262 f| = |256 f| x 2

(262 f) = (256 f) x 2 f = 250, 258 HzUnknown Frequency can not be greater than 262 Hz. because no. of beats heard with 262 Hz is more then

the no. of beats heard with 256 Hz.

|262 f| = |256 f| x 2

(262 f) = (256 f) x 2 f = 250, 258 Hz

vKkr vkofk 262 Hz ls T;knk ugh gks ldrh gS] D;ksafd 262 Hz ds lkFk lqus x;s foLianks dh la[;k 256Hz ds lkFk lqusx;s foLianks dh la[;k ls T;knk gSA

7.4 sin 2nt + sin 2 (n 1) t + sin 2 (n + 1) t

sin 2nt + 2 sin2

)]1n1n(t2[ cos

2

)]1n1n(t2[

sin 2nt + 2 sin 2nt cos 2 t

sin 2nt [1 + 2 cos 2 t] fbeat izLian

= 1

7.5

1=

2

V

2=

4

V

no. of beat heard lqus x;s foLianks dh la[;k n1 n

2=

4

V= 4

if length of pipes are doubled. no of beats heard 1n 2n = 8

V=

2

4= 2

;fn ikbi dh yEckbZ nqxquh dj nh tkrh gS rks lqus x;s foLianks dh la[;k n 2n = 8V =

24 = 2

8.3 n = rs nV

vV

, n

r=

sVV

Vn

n = nVV

VV

s

s

=

50350

50350 1.2 =

300

400 1.2 = 1.6 KHz

8.7

n =

cosVV

sinVV

s

on tan =

2

1is constant and n remains constant and n < n.

tan =2

1rFkk nfu;r jgrs gS vkSj n < n

so. graph. must be

vr% xzkQ ,slk gksxkA

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EXERCISE # 2

PART- I

1.

pipeclosedfor

4

v

pipeopenfor2

v

f .fun

f T , but f does not depend on pressure.for closed pipe f

1st overtone= 3f

fundamental.

fy,dscUn ikbZi

fy,dsikbZi[kqys

4

v2

v

f .fun

f T , ysfdu fnkc ij fuHkZj ugh djrk gSA

cUn ikbZi ds fy, fizFke vf/kLojd = 3fewyHkwr

4. Path difference (S) between direct and reflected wave = 130 120 = 10 m

for so constructive interference S = n

=n

10

n

S

(n = 1, 2, 3, .......)

= 10, 4

10,

3

10,

2

10

..... Ans (A)lh/kh vkSj ijkofrZr rjaxksa ds e/; iFkkUrj (S) = 130 120 = 10 m

vr% lEiks"kh O;frdj.k ds fy, S = n

=n

10

n

S

(n = 1, 2, 3, .......)

= 10,4

10,

3

10,

2

10..... Ans (A)

5. frequency of two source n1

= 50 n2

= 51

nks lzk srks ds e/; vkofr n1

= 50 n2

= 51

so beat frequency = 1/sec.

vr% foLian vkofr = 1/sec.

Now intensity ratio of maximum & minimum value = 221

221

min

max

)aa(

)aa(

=1

4

8

162

vc vf/kdre vkSj U;wure ekuks dk vuqikr = 221

221

)aa(

)aa(

wureU;

vf/kdre

= 1

4

8

162

So Ans Dvr% fodYi D lgh gSA

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6. y1

= A cos (ax + bt)

so here vr% ;gka k = a

2

= a a

2

= b n = 2b

2

reflected

incident

reflected

incident

a

a

64.0

a

reflected= 0.8 A

2

a

a

64.0

ijkorhZr

vkifrr

ijkorhZr

vkifrra

ijkorhZr= 0.8 A

For reflected wave Displacement equation

ijkorhZr rjax ds fy;s foLFkkiu lehdj.ky

2= 0.8 A cos ( ax + bt + )

= 0.8 A cos (ax bt)

by super position law resultant wave

v/;kjksi.k fl)kUr ls ij.kkeh rjaxy = y

1+ y

2

= A cos (ax + bt) + [ 0.8 A cos (ax bt)]

so incorrect option is D Ans D

vr% lgh fodYi D gSA Ans D

7. Due to Doppler effect apparent frequency of S1will continuously decreases. But apparent frequency of S

2

changes to lower value when it crosses o so best represented graph is.Ans (C)

MkWIyj izHkko ds dkj.k S1dh vkofr yxkrkj ?kVrh gSA ysfdu S

2dh vkHkklh vkofr tc ;g o dks ikj djrh gS rks NksVs

ekuks ls ifjofrZr gks tkrh gS vr% lgh xzkQC gksxkA

8. To get beat frequency 1, 2, 3, 5, 7, 8, it is possible when other three tuning fork have frequenci es 551,553, 558, etc.

1, 2, 3, 5, 7, 8, foLian vkofr izkIr djus ds fy;s ;g lEHko gS fd vU; rhuks Lofj=k dh vkofr 551, 553, 558 bR;kfngksA

9. f1

= f0

trainsound

sound

vv

v

= 2.11

f0

1.2 vsound

= vsound

+ v v =5

vsound

f2

= f0

sound

mansound

v

vv = 0.8

2

0

f

f= 1.25

f1

= f0

jay/ofu

/ofu

vv

v

= 2.1

1f0

1.2 v/ofu

= v/ofu

+ v v =5

v /ofu

f2

= f0

/ofu

fDrO;/ofu

v

vv = 0.8

2

0

f

f= 1.25

10. Let original frequency is f ekuk okLrfod vkofr fgSA

by the concept of Doppler effect MkWIyj izHkko dh vo/kkj.kk lsfrequency of reflected wave ijkorhZr rjax dh vkofr

f = f12332

12332f

uV

uV

, f f = 6 ,

320

344f f = 6 f =

24

6320 = 80 Hz

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18.

f = f0

sourcesound

sound

vv

v

/ofu/ofu

/ofu

VV

V=

cosv330330

.600 =37cos20330

330.600

= 600 .

314

330 ~ 630.5 Hz.

19. There is no relative motion between source and observer so frequency remain constant n =0

V

;gka L=kksr vkSj izs{kd ds e/; lkis{k xfr ugh gks jgh gS vr% vkofr fu;r jgsxhA n =0

V

when wind start blowing in the direction of wave motion then velocity of sound = V + uw

tc gok rjax dh fn'kk esa pyuk izkjEHk gksxh rc /ofu dk osx = V + uw

so apparent wave length ' =n

uV w=

V

uV w

0

vr% vkHkklh rjaxnS/;Z ' =n

uV w=

V

uV w

0

Ans (B)

20. frequency of sound for approaching observer fa

=c

vc f

ikl vkus okys iz{ksd ds fy;s /ofu vkofr fa

=c

vc f

For receding observer fr=

c

vcf

nwj tkus okys izs{kd ds fy;s fr = cvc

f

fr+ f

a= 2f f =

2

ff ar

25. Let the velocity of source at mean position is u observer hear maximum

frequency when source approaching line f rom mean positions

ekuk ek/; fLFkfr ij L=kksr dh vkofr u gS rFkk izs{kd vf/kdre vkofr lqusxk tc L=kksr ek/; fLFkrh dh js[kk esa igqaprk gSA

Lmax

=uc

c

and minimum frequency when source reseeding from mean position Lmax

=

uc

c

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vkSj tc L=kksr ek/; fLFkrh ls n wj tkrk gS rc U;wure vkofr Lmax

=uc

c

velocity at mean position ek/; fLFkrh ij osx u = )cos1(gd2

Lmaxvf/kdre = )cos1(gd2c

c

Lmin U;wure

=)cos1(gd2c

c

Ans (C)

26.

AT B Path difference is O and at A path di fference is 4 . From n formula there are 3 maxima positionbetween A & B . So total maxima in ellipse = 16

B ij iFkkUrj O gS vkSj A ij iFkkUrj 4 gSA n fl)kUr ls A vkSj B ds e/; rhu mPpre fLFkrh jgsxhA vr% nh?k Zor

ij dqy vf/kdre= 16 gksxkNote if there were circle, rectangle, square instead of ellipse, answer is same.

uksV ;fn ;gka nh?k Zor ds LFkku ij ok] vk;r ;k ox Z gksrk rks Hkh mRrj leku gk srk gSA

33. Second overtone of open pipe =12

V3

second overtone of closed pipe =24

V5

Since, these frequency are same

21 4

V5

2

V3

52

34

2

1

=5

6

Now, the ratio of fundamental frequencies :

2

1

2

1

4

V

2

V

1

22

= 10 : 6 = 5 : 3 Ans.

gy. [kqys ikbZi dk f}rh; vf/kLojd =12

V3

can ikbi dk f}rh; vf/kLojd =24

V5

pwafd vkofr;ka leku gS

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21 4

V5

2

V3

52

34

2

1

=5

6

vc, ewy fo/kkvks dk vuqikr:

2

1

2

1

4

V

2

V

1

22

= 10 : 6 = 5 : 3 Ans.

40. For stable interference, phase difference should not vary with time.

Hence waves should have same frequency and a constant phase difference.

LFkk;h O;fDrj.k ds fy, dykUrj le; ds lkFk ifjofrZr ugh gksuk pkfg,vr% rjax dh vkofr leku ,oa dkykUrj vpj j[kuk pkfg,A

41.

Wave emitted from Q ls mRlftZr rjax y =A sin (t kxQ)

Wave emitted from P ls mRlftZr rjax y = A sin (t kxp

+2

)

= p

Q

= t kxp

+2

(t kx

Q)

= k (xQ

xp) +

2

=2

(xQ

xp) +

2

=10

(x

Q x

p) +

2

For A ds fy, xQ

xP= 5

=10

( 5) +

2

= 0

For B ds fy, xQ

xP

= 5

= For C ds fy, x

Q x

P= 0

=2

C= (+ + 2cos0) : (+ + 2cos) : (+ + 2cos

2

)

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42. When sound wave is reflected from rigid end displacement wave get extra phase at and pressure wave get noextra phase

So option A, B, C are correct .

tc /ofu rjaxs n

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5.time taken by sound waves to travel dx

/ofu rjax }kjk dx nwjh r; djus esa fy;k x;k le;

dt =V

dx=

M

RT

dx

=

RT

M

dx

dt =

1

M

R

T

dxT = T

1+ x

d

TT 12

dt =

1

M

R

xd

TTT

dx

121

t =

1

M

R

12 TT

d2

d

121 x

d

TTT

=12 TT

d2

1

M

R

12 TT =

12 TT

d2

1

M

R

Vo=

M

RTo(given) fn;k gS

o

o

T

V=

M

R

t =12 TT

d2

1

o

o

T

V

= 21 TT

d2

o

o

V

T=

oV

d2 21

o

TT

T

=

360

9002

900100

400

= 2.5 sec

6. Assume a spherical surface of radius a passing through the ring. The waves passing through ring also passthrough sphere.

oy; ls ikl gksrh gqbZa

f=kT;k dh ,d xksyh; lrg dh dYiuk djrs gSA rjaxs oy; ds lkFk&lkFk xksys ls Hkh xqtjrh gSAarea of ds = 2a sinadds dk {ks=kQy = 2a sinad

s = 2a2

0

dsin

s = 2a2 [cos + cos 0]s = 2a2 [1 cos]

Energy tkZ = sa4

power2

sa4 2

kfDr'=

2

power

2

kfDr'(1cos)

= 24

power

2

4

kfDr'

Energy tkZ =2

l4 02

(1cos)

< > = 220

2)/R(1

11

= 2l20

22 5.0l

l1

w20 Ans

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EXERCISE # 3

1.1 (A) y = 4 sin (5x 4t) + 3 cos (4t 5x + /6)

is super position of two coherent waves moving in positive direction, so their equivalent will be an another

travelling wave.

(B) y = 10 cos

330

xt sin (100)

330

xt lets check at any point, say at x = 0,

y = (10 cos t) sin (100 t)

at any point ampli tude is changing sinusoidally. so this is equation of beats.

(C) y = 10 sin (2x 120t) + 10 cos (120t + 2x) = superposition of two coherent waves travellingin opposite direction.

equation of standing waves.(d) y = 10 sin (2x 120 t) + 8 cos (118t 59/30x) = superposition of two waves whose frequencies areslightly different (

1= 120,

2= 118)

equation of Beats.(a) y = 4 sin (5x 4

t) + 3 cos (4t 5x + /6)

nks /kukRed fn'kk esa xfreku dyk lEc) rajxks dk v/;kjksi.k gS] bl izdkj budk rqY; Hkh ,d vU; izxkeh rjax gSA

(b) y = 10 cos

330

xt sin (100)

330

xt fdlh fcUnq ij tkp djrs gS] x = 0 ij,

y = (10 cos t) sin (100 t)

fdlh Hkh fcUnq ij vk;ke T;koh; ifjofrZr gksrk gSA bl izdkj ;g foLiUnksa dh lehdj.k gS(c) y = 10 sin (2x 120t) + 10 cos (120t + 2x) = foijhr fn'kk esa lapkfjr nks dyk lEcU/k rjaxks dk

v/;kjksi.kvizxkeh rjxksa dk lehdj.k(d) y = 10 sin (2x 120 t) + 8 cos (118t 59/30x) = nks rjax ks dk v/;kjksi.k ftudh vko fr;ksa esa tjk lk vUrjgS (

1= 120,

2= 118)

foLiUn dh lehdj.k

1.2 Find all speeds w.r.t. wind and apply doppler effect .

lHkh pkyksa dks gok ds lkis{k ysus ij vkSj MkWIyj izHkko dk mi;ksx djus ij(A) The wavelength of the waves coming towards the observer from source

L=kksr ls izs{kd dh rjQ vkrh gqbZ rjax dh rjaxnS/;Z= (V - V

W+ V

S)/f

(B) The wavelength of the waves incident on the wall

nhokj ij vkifrr gksrh gqbZ rjax dh rjaxnS/;Z= (V + V

W- V

S)/f

(C) The wavelength of the waves coming towards observer f rom the wall = (V VW V

D)/f

r

nhokj ls izs{kd dh rjQ vk jgh rjax dh rjaxnS/;Z(D) Frequency of the waves (as detected by O) coming from wall after reflection = (V - V

W- V

O)f

r/(V - V

W-

VD)

tc nhokj ls ijkorZu ds ckn rjax dh vkofk izs{kdO ds }kjk izsf{kr)

= (V - V

W- V

O)f

r/(V - V

W- V

D)

2.1 = (0.1 mm) cos80

2(y + 1

cm) cos 2(400) t

end correction is 1cm. ,so at y = -1 cm.

= (0.1 mm) cos80

2(1 cm+ 1

cm) = (0.1 mm) cos (0) = Antinode

So upper end is open.

at lower end y = 99 cm

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= (0.1 mm) cos80

2(99 + 1)

= 0.01 cos2

5= 0 Node

tube is closed at lower end

So tube is open closed.

= (0.1 mm) cos80

2(y + 1

cm) cos 2(400) t

fljk la'kks/ku 1cm. ,vr% y = 1 cm ij

= (0.1 mm) cos80

2(1 cm+ 1

cm) = (0.1 mm) cos 2(0) = izLiUn

blfy, ijh fljk [kqyk gqvk gSuhpys fljsa ij y = 99 cm = 0.99 m

= (0.1 mm) cos80

2(99 + 1)

= 0.01 cos2

5 = 0 fuLiUn

uyh fupys fljs ij cUn gSAblfy, uyh [kqyh cUn gS

2.2

2

80

2So = 80

and effective length of air column = 99 + 1 = 100 cm

So4

5

l= 5

4

, so five half loops will be formed

open

closed

= 5

4

so second overtone.

2

80

2 vr% = 80

ok;q LrEHk dh izHkkoh yEckbZ = 99 + 1 = 100 cm

blfy,4

5

l = 5

4

, blfy, ikp v)Z ywi cusxs

Open ( )[kqyk

Closed ( )cUn = 5 4

blfy, nwljk vf/kLojd

2.3 Pex

= Bdx

d

= (5 105) (0.1 103)80

2sin

80

2(y + 1cm) cos 2(400) t

= (125 N/m2) sin80

2(y + 1cm) cos 2(400t)

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4.1 I = 2r4

P

211

211

21

22

2

1

2

1

rP2

)r2(P

r

r

P

P

I2

=2

2.0

2

1

= 0.1 w/m2.

4.2 =V2

P2m

1012 =(300))5.1(2

P2m

P = 3 10 Pm a-5

Pm

= BAK

= (v2) A

f

v

2

Pm

= (2Vf)A

3 10-5 = (2) (1.5) (300) (103) A

A = 1/3 10 m -10

4.3 for stationary wave vizxkeh rjax ds fy;s Y = y1

+ y2

2a sinkx cost = a sin(tkx) + y2

y2 = a sin (k x + t) + a sin (k x t) a sin(tkx)y

2= a [sin (kx +t) + 2sin(kx t)] Ans

4.4 effective length izHkkoh yEckbZ = 83.2 + 0.3d = 85 cm

n =85.04

340

4

V

= 100 (fundamental frequency ewy vkofr )

other resonating frequency below 1000 Hz

1000 Hz ls uhps dh vU; vu quknh vkofr;kW100, 300, 500, 700, 900

so four overtone below 1000 Hz.

vr% 1000 Hz ls uhps 4 vf/kLojd

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EXERCISE # 4PART-I1. The motorcyclist observes no beats. So the apparent frequency observed by him from the two sources must be

equal.

f1= Frequency recorded by motorcyclist from police car.

f2 = Frequency recorded by motorcyclist from stationary siren.For no beats f

1= f

2

176

330

330165

22330

330

Solving this equation we get, = 22 m/seksVj lkbZfdy pkyd dksbZ foLian ugha lqukrk gS] blfy, mlds }kjk lquh xbZ nksuksa lzksr dh vkHkklh vkofk cjkcj gksuh pkfg,Af1= iqfyl dkj dh eksVj lkbZfdy pkyd }kjk lquh xbZ vkofk

f2= :dh gqbZ dkj dh eksVj lkbZfdy pkyd }kjk lquh xbZ vkofk

dksbZ foLian ugha lquus ds fy, f1= f

2

176

330

330165

22330

330

gy djus ij ge izkIr djrs gSa , = 22 m/s

2. Let e be the end correction.

0.1 + e =4

0.35 + e =

4

3

Solving this equation we get

e = 0.025 m = 2.5 cm

ekuk e fljk la'kks/ku gSA fn;k x;k gS ,

0.1 + e = 4

0.35 + e = 4

3

gy djus ij ge izkIr djrs gSa e = 0.025 m = 2.5 cm

3. Fundamental frequency, f=)r6.0(4

v

speed of sound v = 4f ( + 0.6r)or v = (4) (480) [(0.16) + (0.6)(0.025)] = 336 m/s

0.6r

ewy vkofrZ , f =)r6.0(4

v

/ofu dh pky v = 4f ( + 0.6r);k v = (4) (480) [(0.16) + (0.6)(0.025)] = 336 m/s

4. The frequency is a characteristic of source. It is independent of the medium. Hence, the correct option is (D).

vko fk, L=kksr dk vfHkyk{kf.kd xq.k gSA ;g ek/;e ls Lora=k gSA vr% lgh fodYi (D) gSA

5. fc= f

0(both first overtone) (nksuksa izFke vf/kLojd)

or;k

L4

v3 c =

0

0

L2

v2 L

0= L

v

v

3

4

c

0

= L

3

4

2

1

as D;ksfd v

1.

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7. 1.8 103 = fV300

300

s

2.2 10

3 = fV300

300

s V

s= 30 m/s

8. 100f2 1 f1 = 502f

2= 92 f

2= 46

beat frq = 50 46 = 4foLian vkofk = 50 46 = 4

9. v =k

=

5.0

100

Speed pky v = 200

10. At x = 0,

y = y1

+ y2

= 2A cos 96 t cos 4 tFor y = 0 ds fy;s, cos 96 t = 0 or;k cos 4 t = 0

for 0 < t < 1 cos 4 t become 0 at t =8

1,

8

3,

8

5,

8

7sec

0 < t < 1 ds fy, cos 4 t, le; t =8

1,

8

3,

8

5,

8

7 lSd.M ij 'kwU; gksxk

but cos 96 t is not 0 at these values but cos 96 t become 0, 96 times in 1 sec.Therefore net amplitude become zero 100 times.

ysfdu cos 96 t bu ekuks ds fy, 'kwU; ugh gS ysfdu cos 96 t, 1 lSd.M esa 96 ckj 'kwU; gksxkvr% dqy vk;ke100 ckj 'kwU; gksrk gSA

11. VS/A

= 340 + 20 = 360

20

A B

340340

20 30

VS/B

= 340 30 = 310

12. Because

VS = VO = 20so as seen from passengers of train A,

f = f0

frequencyf = f1 1 f = f2 2

Intensity

13. As seen from B

fmin

= )800(20340

30340

)1120(20340

30340fmax

310ff minmax

VS/A = 340 + 20 = 360

20

A B

340340

20 30

VS/B

= 340 30 = 310

D;ksafdV

S= V

O= 20

Vsu A ds ;k=kh }kjk ns[kk x;k ,f = f

0

B }kjk ns[kk x;k A

fmin

= )800(20340

30340

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)1120(20340

30340fmax

310ff wureU;vf/kdre

14. As string and tube are in resonance f1

= f2

|f1 n| = 4 Hz.When T increases, f

1also increases. It is given

that beat frequency decreases to 2 Hz.

n f1

= 4

n = 4 + f1

as f1= f

2

n = 4 + f2

f2

=4

V3=

)4/3(4

3403

= 340

n = 344

pwafd jLlh rFkk V~;wc vuqukn esa gSA f1

= f2

|f1 n| = 4 Hz.tc T c

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f =)vv)(vv(

f)vv(v2

21

012

v

f)vv(2 012

Givenv

f)vv(2 012 =

100

2.1f0

v2 v1 = 7.126Answer in nearest integer is 7.

ekuk dkjksa dh pky V1o V2gSA dkj }kjk izkIr vkoZrh f1 =

1vvv

f0

dkj }kjk ijkofrZr vkoZrh f2 =

v

vv 1

1vvv

f0

f = f2 f2 =

1

1

2

2

vv

vv

vv

vvf0

f =

)vv)(vv(

)vv)(vv()vv)(vv(

21

2112f0

f =)vv)(vv(

f)vv(v2

21

012

v

f)vv(2 012

fn;k x;k gSv

f)vv(2 012 =

100

2.1f0

v2 v1 = 7.126fudVre iw.kkZd esa mkj 7 gSA

17. Fundamental frequency of close organ pipe =1

1

4

V

cUn vuqukn uyh dh ewy vkofk =1

1

4

V

Second harmonic frequency of string =2

2

2

V2

jLlh dh f}rh; lauknh vkofk =2

2

2

V2

So vr%,1

1

4

V

=

2

2V

=8.04

320

=

5.0

1

50

2500 =50

=50

1=

5.0

m

m = 10 gm.

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20.

Open end will act like pressure node & close end will act l ike pressure antinode so for reflected wave will be high

pressure for close end and will be low pressure for open end.

Ans. B & D

21.

First resonance occurs at fundamental frequency

f =V

4( e) + e =

V

4f(where e = 0.6 2 = 1.2 cm)

+ e =5124

336

= 0.164 m

= 16.4 1.2 = 15.2 cm

PART - II

1. Let the tubes A and B have equal length called as l. Since, tube A is opened at both the ends, therefore, itsfundamental frequency

ekuk nksuks uyh;ksa A rFkk B dh yEckbZ leku () gSA D;ksfd A, nksuks fljks ls [k qyh gS vr% ewykofr

nA = 2

v

Since, tube B is closed at one end, therefore, its fundamental frequency

D;ksfd B ,d fljs ls cUn gS blfy;s bldh ewykofrZ

nB = 4

v

From eqs. (1) and (2), we get

lehdj.k (1) vksj (2) ls ge izkIr djrs gS fd

B

A

n

n=

4/v

2/v

=2

4= 2 : 1

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////

///

/////

//

/////DS

2. The tuning fork of frequency 288 Hz is producing 4 beats /sec with the unknown tuning fork i.e., the frequencydifference between them is 4. Therefore, the frequency of unknown tuning fork= 288 4= 292 or 284On placing a little wax on unknown tuning fork, its frequency decreases but now the number of beats producedper second is 2 i.e. the f requency difference now decreases. It is possible only when before placing the wax, the

frequency of unknown fork is greater than the frequency of given tuning fork. Hence, the frequency of unknowntuning fork = 292 Hz

288 Hz vkofr dk Lofj=k vKkr Lofj=k ds lkFk 4 foLian/lSd.M mRiUu djrk gS vFkkZr~ nksuks ds e/; vkofr vUrjky 4 gSAvr% vKkr Lofj=k dh vkofr= 288 4= 292 or 284

vKkr Lofj=k ij FkksMk lk ekse yxkus ij mldh vkofr ?kVrh gS ysfdu vc izfr lSd.M foLianks dh la[;k 2 gksxh A vFkkZr~vkofr vUrjky ?kVrk gSA og rHkh lEHko gS tc] vKkr Lofj=k dh vkofr, ekse yxkus ls igys, fn;s x;s Lofj=k dh vkofr lsvf/kd gksxhAvr% vKkr Lofj=k dh vkofr = 292 Hz

3. Frequency of tuning fork decreases with temperature.

rkieku ds lkFk Lof j=k dh vkofk ?kVrh gSA4. Frequency of string is 256 5.

Since number of beats is decreasing when frequency of string is increasing so frequency of string is 256 5.

jLlh dh vkofk 256 5 gksxhtc jLlh dh vkofk c

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8. 1 = 10 log0

1

2 = 10 log

0

2

1 2 = 10 log2

1

= 20

log2

1

= 2

2

1

= 100

9. V02=

M

RT=

32

RT

5

7= 460

VHe=

M

RT=

4

RT3

5

=7

532460460

12

5 = 1419 m/s

10. During summer speed of sound increases. So wavelength increases.

so x > 3 18 so x > 54

xfeZ;ksa esa /ofu dh pky c 54

11. Motor cycle has travelled a distance s. Its velocity at that point

v = as2

The observed frequency

f' = f330

v330

0.94 = 330v330

v = 0.06 330 m/s= 19.8 m/s

s =22

8.19

a2

v 22

= 9.92 = 98 m

eksVj lkbZfdy s nwjh r; djrh gSA ml fcUnq ij bldk osx

v = as2

sf{kr vkofk

f' = f330

v330

0.94 = 330v330

v = 0.06 330 m/s= 19.8 m/s

s =22

8.19

a2

v 22

= 9.92 = 98 m

12.

7. sound wave

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