Class 11-science H C VERMA Solutions Physics Chapter 16 - Sound Waves

Time gap = t_air - t_steel

= d/v_air - d/v_steel

=1/330 - 1/5200

= 2.75 × 10^-3 sec

Δt = 2.75 ms

Distance travelled by sound to come back = 80×2 = 160 m

Time interval = distance/speed

= 160/320

= 0.5 sec

Time gap between two claps = 3/10 =0.3 sec

Distance travelled by sound between two claps = 50+50

= 100 m

Velocity = distance/time

= 100/0.3

V≈333 m/s

v = fxλ

For λ_max = v/f_min

=360/20

=18m

For λ_min = v/f_max

=360/20×10³

 λ_min = 18 mm

V=f × λ 

For λ _max  = V/f_min

=1450/20

λ _max = 72.5 m

For λ _min = V/f_max

=1450/20000

λ _min = 7.25 cm 

(a)λ_sound = 10.d_speaker

=10(20)

=200 cm=2m

V=f ×λ 

f=340/2 = 170 Hz

(b) λ_sound = d/10

=20/10

=2cm

=0.02 m

V= f ×λ 

F=340/0.02

=17000 Hz

V=f×λ

In air

λ =v/f

λ _air = 340/4.5×10⁶

λ _air=7.6×10^-5m

In tissue

λ =v/f

λ _issue = 1.5×10³/4.5×10⁶

λ _t = 3.3×10^-4m

y=6sin(600t-1.8×)

Comparing equation with y=Asin (wt±Kx)

K=1.8; w=600; A=6×10^-5m

2/ λ =1.8

λ=2 /1.8

(a)

A/λ = 6×10^-5/(2 /1.8)=1.7×10^-5

(b)

Wave speed = (coefficient of t)/(coefficient of x)

V_w=600/1.8 = 100/3 m/s

Velocity Amplitude = A 

=(6×10^-5)(600)

V_p=36×10^-3m/sec

Ratio=V_p/V_w=36×10^-3/(100/3) = 1.1×10^-4

V=f×λ

350=100×λ 

λ=3.5m

(a)

△Ø=.t

=(2 f)t

=(2)(100)(2.5×10^-3)

= /2

(b)

△ Ø=. △× 

=.x(0.1)

=2 /35

(a)Path difference of two waves = 30-20

△=10 cm

Ø = .△ 

=×10

Ø =4 

Since Ø =2n i.e., constructive interference

So phase difference is zero

(b) Path difference = 0

So, Ø=0

Density= mass/volume

= 

Adiabetic Constant = C_p/C_v

=3.5R/2.5R

v= 

==310m/s

v 

V₁/V₂= 

340/V₂= 

V₂=349m/s

V 

V₁²/V₂²=T₁/T₂

v₁²/(2v₁)²=(0+273)/T₂ [∵v₂=2v₁]

T₂=273×4K

T₂=273×4-273

T₂=819°C

The variation of temperature at a distance x from surface of T₁ temperature

T=  .x+T₁

Velocity of sound V 

= 

V_T==v 

= x V

⁼  

t =  

Putting T₁=280K, T₂=310K,d=33m and v=330m/s

t= 96 ms

v= 

=v² 

= v² 

△V= 

 =  

△V=0.14 cm³

P=βAK

β= 

β=/△ 

β=1.4×10⁵N/m²

(a) I=P/A=P/4r²

I= 

I=44 mW/m²

(b) I=(pressure amp.)²/2 

P_max= 

₌ 

P_max=6 N/M²

(c) I=2²f²A²v

A= 

A=1.2* m

I 

I₁/I₂=r₂²/r₁²

10^-8/I₂=(25)²/(5)²

I₂=4×10^-10w/m²

△L=10log 

=10log 

=10log(5/50)²

40-L=-20dB

L=20dB

△L=10log 

=10log 

△L=3dB

I=P/A=P/4r²

r= 

=  [∵ when L=120dB then I=1w/m²]

r=0.4m=40cm

△L=10log 

(60-50)=10log 

=10

Iρ²

=√10

△L=10log 

Let I be the intensity produced by each student

△L=10log 

L-50=10×0.3

L=53dB

Path difference between consecutive maxima and minima=/2

If tube is slided by x, then

2×= /2

 =2×(2.5*10^-2)*2

 =10^-1m

V=f* 

f=340/10^-1

f=3.4kHz 

(a)

Path difference between consecutive maxima and minima= /2

If tube is slided by x

2x= /2

 =4×16.5×10^-3

 =66×10^-3m

V=f* 

330=f×66×10^-3

f=5kHz

(b)

I_max/I_min=(A₁+A₂)²/(A₁-A₂)²

9I/I=(A₁+A₂)²/(A₁-A₂)²

A₁/A₂=2

Path difference of the sound waves = 6.4-6

△=0.4m

For destructive interference

△= 

0.4= 

f= × (2n+1)

f=400(2n+1)

Putting n=1,2,3… till f˂5000Hz

So, frequencies are 1200 Hz, 2000Hz, 2800Hz, 3600Hz and 4400Hz

If board is moved by x then extra path difference between waves is 2x.

Path difference between consecutive maxima and minima is λ/2

So, λ/2=2x

λ =4×20×10^-2

λ =0.8m  

v=f× λ 

f=336/0.8

f=420Hz

Initially, path difference is =2[ ]- d

△i=2d

Let cardboard is shifted by distance x, then final path difference = 2[ ]- d

Path difference between consecutive maxima and minima is λ/2 =  

∆_net=∆_f-∆_i 

d/4 = 2[-d ]- 2d

On solving

X=0.13d

Path difference =  - 3.2

△=0.8m

For destructive interference,

△= = 0.8

(2n+1)(320/f) = 2×0.8

f=200(2n+1)

Lowest frequency heard when n=0 will be f=200Hz

For Highest frequency f=20,000Hz=200(2n+1)

N=49.5

So maximum value of n is 49

On putting n=0,1,2,…..49 person will hear in audible range.

Let detector is moved by x at point P

Path difference between consecutive maxima and minima is λ/2 = 20/2 = 10cm

△=S₂P - S₁P

10= - 

On solving,

X=12.6cm

V=fxλ

λ=330/600 = 11/20 m

(a)

D-1 will be observed here

△=dsinθ 

λ/2 = dsinθ

½ x 11/20 = 2sinθ 

sinƟ = 11/8

Ɵ=11/80 rad. [∵ angle is small sin≈Ɵ]

Ɵ=11/80×180°/π=7.9°

(b)

B-1 will be formed

△=dsinƟ

λ =dsinƟ

11/20 = 2sinƟ 

sinƟ=11/40

Ɵ=11/40x180°/π=16°

(c)

Let n^th be last maxima obtained at θ=90° 

△=dsinƟ

nλ=dsin90°

N(11/20) = 2

n=3.63

So, last maxima is for n=3.

So, he get two maxima from point P along the line as he walks. 

∵ IA²

∵A₁=A₂=A₃=A

By vector method, 

Resultant Amplitude =0

So,

I_N=0

At point O,

△=S₁O-S₂O

△=2λ(constructive interference B-2 order)

So,

At point P, B-1 will be formed

△=dcosƟ 

1×λ=2λcosƟ 

Ɵ=60°

In △S₁PO

tan60°=x/D

X=√3D

∆=dcos Ɵ 

Ɵ =cos^-1(△/3 λ)

B-0; △=0 => Ɵ=90° 

B-1; △=λ => Ɵ =70.5° 

B-2; △=2 λ => Ɵ =48.2° 

B-3; △=3 λ => Ɵ =0° 

Let intensity of each source be I

(a)

 At P

∆=S₂P-S₁P=0

So, Ø=0 (constructive interference)

I_max=(√I₁+√I₂)² = (√I+√I)²

I_o=4I

Now, when one source is switched off intensity at this point will be I i.e., I_o/4

(b)

If Ɵ=60° then also △=0

So, no change in answer

Fundamental frequency of O.O.P =V/2l

=340/2(0.2)

=850Hz

I^st overtone frequency =2f

f₁=2(850)

f₁=1700Hz

II^nd overtone frequency = 3f

f₂=3(850)

f₂=2550Hz

Fundamental frequency of C.O.P.=V/4l

500=(340)/4l

l=0.17m=17cm

Distance between consecutive nodes = λ/2 = 4cm

V=f×λ

328=f×8/100

f=4.1KHz

Distance between consecutive node and antinode is λ/4

λ/4=25cm

λ=1m

v=fxλ

f=340Hz

Frequency in O.O.P.

f=nv/2l

f=n(340)/2(0.5)

f=n(340)

So, frequencies between 1000Hz and 2000Hz will be when n=3,4,5 i.e. f=1020,1360,1700Hz respectively.

L₂-L₁= λ /2

62-20= λ /2

λ =84cm=0.84m

(a)

V=fx λ 

V=400(0.84)

V=336m/s

(b)  

L₁+e= λ /4

20+e=84/4

e=1cm

I_st overtone frequency of cop=fundamental frequency of oop

3×v/4l_c = v/2l_o

3/(4×30) = 1/2×l_o

L_o=20cm

For fundamental frequency

λ/2=1

λ=2m

v=f×λ

3800=f×2

f=1.9kHz

Other frequency are given by = n×1.9KHz

where n=1,2,…. 10 in range of 20Hz-20,000Hz

f=v/2l

L_max=v/2f_min

L_max=340/(2)(20) = 8.5m

(a)

Fundamental frequency of o.o.p = v/2l

=340/2(0.05)

f=3.4KHz

(b)

Let n^th harmonic be highest in audible range

f^'=nf

20,000=n(3400)

n=5.8

So,n=5

Fundamental frequency f cop== =100Hz

Higher harmonics frequencies = (2n+1)100

Where n=0,1,…….. 9

Difference between two successive resonance frequency in O.O.P. = - =   

(2592-1944) =  

l=0.25m

l=25cm

Path difference between two consecutive maxima = λ=2×32 cm

λ=0.64m

v=f×λ

=512(0.64)

v=328m/s

Fundamental frequency in shorter arm for cop= 

440= 

=18.8cm

In longer arm, I^st overtone frequency =  

440= 

=0.563m or =56.3cm

II^nd harmonic of wire = fundament frequency of cop

 =  

 =  

 [∵ =mass/length]

T=11.6N

Fundamental frequency of wire = fundamental frequency of cop

 =  

 =  

T=347N

f  

f  

by error

 =  

f  

=294Hz

_Rod/4 = 25cm= 100cm=1m

λ_air/2 = 5cm =10cm=0.1m

v_air/λ_air=v_rod/λ_red 

340/0.1 = v_rod/1

v_rod=3400m/s

L_rod=λ_r/2 

λ_r =2m

λ_gas/2=6.5cm

 λ_gas=13cm

f=v_air/λ_a=v_rod/λ_r 

2600=v_air/13×10^-2=v_rod/2

V_air=338m/s and v_rod=5200m/s 

Frequency of source is 476Hz or 480Hz

Number of beats is 2

So, possible frequency of tuning fork=476±2 or 480 ±2

=474,478 or 478,482

So, common frequency of tuning fork=478Hz

Frequency of tuning fork 'A'=256Hz

No. of beats=4

So, frequency of tuning fork 'B'=256±4

'B'=252Hz or 260Hz

Now, on waxing fork 'B', its frequency will decrease and will produce and beats with A which is possible when frequency of fork B is 252 Hz

Beat frequency = f₂-f₁

=v/λ₂-v/λ₁ 

=- 

≈7Hz

Fundamental frequency of cop=v/4l

=(320)/[(4)(0.4)]

=200Hz

Tuning fork produces 5 beats with cop

f_fork=205 or 195Hz

When loaded with wax, frequency will decrease.

∵Beat frequency decreases

f_fork=205Hz

Initially,

F_A=f_B=600Hz

∵f√T

As tension increases, frequency increases

So, f_A=606Hz; f_B=600Hz

f_A/f_B=√T_A/√T_B

606/600 = √T_A/√T_B 

T_A/T_B=1.02

Frequency of wire=256±4

=252Hz or 260Hz

f_wire= = f_wire 

By shortening, the length of wire, f_wire increases.

∵Beat frequency decreases

So, f_wire = 252Hz

Initially, f_wire=252Hz and length of wire = 25cm

Later, f_wire=256Hz and length of wire = l_f

f_i/f_f=l_f/l_i

252/256 = l_f/25

l_f=24.6cm

Length shortened = 25-24.6

=0.6cm

V_o=10m/s

f_o=2000Hz

f_app= f_o []= 2000 []

f_app=2.06KHz

v_s=18× = 5m/s

f_o=2400Hz

f_app=f_o[]

2400[]

f_app=2436Hz

(a)  

V_s=20m/s

f_o=1250Hz

f_app= f_o[]

=1250[]

=1328Hz

(b)

f_o=1250Hz

f_app= f_o[]

=₁₂₅₀[]

f_app=1181Hz

Initially,

f_app=f_o[]=1620 ---(1)

F_app=f_o[[]----(2)

Divide(1) and(2),

1620/F_app=v+v_s/v-v_s

1620/F_app=340+15/340-15

F_app=1480Hz

Apparent frequency received by wall Y is

f' = f_o[]

Apparent frequency received by bat after reflection from the wall Y is

f'' = f'[]

=4.5×10⁴ 

f''=4.66×10⁴Hz

Apparent frequency received by wall x is

n'=fo[]

Apparent frequency received by bat after reflection from the wall x is

n''=n'[]

=4.5×10⁴ 

n''=4.33×10⁴Hz

Beat frequency = 4.66×10⁴-4.33×10⁴

=3300Hz

Apparent frequency heard before bullet crosses person

f₁=f_o[]

=f_o[]

=3 f_o

Apparent frequency heard after bullet crosses person

f₂=f_o[]

=f_o[]

F₂=0.6f_o

So, f₂/f₁=0.6f_o /3f_o=0.2

∴Fractional change = 1-0.2

=0.8

Velocity of train = 72kmph

=72× 

=20m/s

Frequency heard by person from train A=f_o[] 

=500[] 

=529Hz

Frequency heard by person from train B = f_o[] 

=500[] 

=474Hz

(a)

Beat frequency for standing man = 4

So, apparent frequency heard by standing man = 440± 4

=444Hz or 430Hz

Since, source is coming towards observer so f_app=444Hz

f_app=[]f_o

444=[].440

V_s=3.09m/s

=3.09×=11kmph

(b)

The sitting man will listen to fewer than 4 beats/sec

f₁=f_o[]

=256[]

=258.3Hz

f₂=f_o[]

=256[]

f₂=253.7Hz

Beat frequency = f₁- f₂

=4.6Hz

f₁=f_o[]

=512[]

=520.6Hz

f₂=f_o[]

F₂=512[]=503.6Hz

∴ Beats = f₁-f₂

=17.5Hz

Velocity of source = Rw ==10m/s

Maximum and Minimum frequency will be observed at point PandQ respectively

f_{app /max}=f_o[] 

=500[]=515.5Hz

f_app/min=f_o[] 

=500[]=485Hz

v_s=90× =25m/s v_o=25m/s

F_o=500Hz

Apparent frequency = f_o[v+v_o/v-v_s]

=500[350+25/350-25]

=577Hz

f_o=16Hz

f_app=20KHz

f_app=f_o[]

20×10³=16× 10³× []

V_o=330/4 m/s

=330/4×18/5=297Kmph

This speed is not practically attainable for ordinary cars

f_o=800Hz

f_app=f_o[(v-v_o)/(v-v_s)]

f_app=800[(330-20)/(330-30)]≈827Hz

(a)

Frequency received by second submarine

F_app=f_o[(v+v_o)/(v-v_s)]

=2000[(1500+15)/(1500-10)]=2034hz

(b)frequency heard by first submarine

F'_app=2034[(1500+10)/(1500-15)]

=2068Hz

Maximum and minimum frequency will be observed at mean position P and Q respectively as velocity of source will be maximum

f_max=f_o[ ; f_min=f_o[ 

f_max- f_min=8

f_o[ - f_o[ = 8

2f_ovv_s/v²-v_s²=8

V_s=1.7m/s

Aw=1.7

W==10

T=2/w=2 /10

T=0.63sec

f_o=1650Hz 

f_app=f_o [ 

=1650[ 

f_app=1670Hz

(a)

f_app=f_o[  

=660[  

=680Hz

(b)

No. relative velocity along line joining source and observer

So, no. Doppler effect f_app=660Hz

(c)

f_app=f_o[  

=660[  

=640Hz

(a)

No relative motion between passenger and train Hence f_obs=500Hz

(b)  

f_o=500Hz

f_app=f_o[ 

=500[ = 459Hz

(c)

When wind starts blowing,

Frequency heard by passenger is unaffected as no relative motion,

Frequency heard by person standing near track

F_app=f_o[  ]

=500[]

=458Hz

(a)

Frequency received by wall

f_app=f_o[  ]

=1600[]

=1616Hz

(b)

Reflected frequency received by boy = f[]

=1616[]

=1632Hz

Frequency received by car

F_app=f_°[]

=1600[]

=1503Hz

Reflected frequency received by boy

f'_app=f_app[]

=1503[]

=1417Hz

Velocity of car = 54× = 15m/s

(a)

Net velocity of wave in front of the car = v_air-v_car=335-15=320m/s

V=f×λ 

320=400×λ 

Λ=0.8m or λ=80cm

(b)

Frequency received by the diff

f_diff=f_o[]

=400[]

=418.75Hz

V=fxλ 

335=418.75×λ 

λ=80cm

(c)

Frequency of the reflected sound wave heard by the person sitting in the car

f_app=f_cliff[]

=418.75()

f_app=437Hz

(d)

Beat frequency =437-400

=37

But a human-being can hear maximum of 10 beats in one second

So, he will not hear beats but continuous sound.

Frequency of sound heard at car

f=f_o[]

f=400[]

Frequency of the reflected sound by observer

f'=f[]

410=400[][]

V_car=4m/s

(a)

Speed=distance/time

330=330/t

t=1sec

(b)

Frequency of sound heard by listener = 2khz

Since, no velocity is along line joining source and observer hence, no Doppler effect

(c)

For source

V_s=d/t

d=22×1

d=22m away from P on x-axis

At t=0, let the source be at a distance of y from origin. Now, time taken by source to reach origin is same as time taken by sound to reach observer

y/22= 

6=660/√224

Frequency heard by the observer

f_app=f_o[  ]

=4000[ ]

f_app≈4018Hz

(a)

Time taken by source to reach intersection point is equal to the time taken by sound to reach detector

y/170 = 

y=200/√3

Frequency of sound heard by detector

f_app=f_o[ ]

 =1200[  

 =1600Hz

(b)

Detector will detect frequency of 1200Hz when sound was produced at intersection point.

Time to by sound to reach detector

T=200/340

In this time source moves by a distance of

V=d/t

D=170× = 100m

So, source-detector.distance =  

=224m

(a)

Maximum frequency observed when maximum velocity of source is towards observer

f_app)A=f_o[]

=500[]

f_app)A≈514Hz

(b)

f_app)D=f_o[]

=500[]

=490Hz

F_app)B=f_o[]

=500[]

=511Hz

Let the distance between the source and observer is x at t=0

Time taken for 1^st pulse to reach observer t₁=x/v

II_nd pulse starts after T(where T=1/f)

and it travels a distance of ()

So, t₂=T+ 

Now, t₂-t₁=(-)-()

t₂-t₁=T- 

Putting T=1/f

t₂-t₁=(2fv-a)/(2fv²)

Beat frequency = 1/t₂-t₁=2fv²/2fv-a