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HC Verma Class XII Physics Chapter 16 – Sound Wave


Exercise : Solution of Questions on page Number : 351


Answer: 1

No, we cannot hear the sound of stones. Sound is a mechanical wave and requires a medium to travel; there is no medium on the moon.
No, we cannot hear the sound of our own footsteps because the vibrations of sound waves from the footsteps must travel through our body to reach our ears. By that time however, the sound waves diminish in magnitude.


Answer: 2

Yes, we can hear ourselves speak. The ear membrane, being a part of our body, vibrates and allows sound to travel through our body.
No, we cannot hear our friend speak as there is no medium (air) through which sound can travel.


Answer: 3

A longitudinal wave propagates when the rod is hit vertically.

When hit horizontally too, a longitudinal wave is produced (sound wave). However, if the rod vibrates, the wave so developed is transverse in nature.


Answer: 4

It depends on the position of the speakers. The placement decides whether the interference so formed is constructive or destructive.


Answer: 5

The frequency of sound produced by vibration of vocal chords is amplified by resonance in the voice box. Now resonant frequency is directly proportional to the velocity of sound present in the voice box. Now as Helium has less density than air, velocity of sound in Helium is higher than that in air. Higher velocity of sound in Helium implies that the resonant frequency of the sound in voice chamber filled with Helium will be higher than with air. Thus the voice is high pitched in Helium filled voice box.


Answer: 6

The displacement node is a pressure anti-node and via-versa.


Answer: 7

We know that: intensity ∝ (amplitude)2.
However, the intensity is independent of frequency. As the amplitude of the vibrating forks is the same, both the forks produce sounds of the same intensity in the air.


Answer: 8

The frequency of the sound is still that of the source. However, the frequency of the vibrations received by the observer changes due to relative motion.
If both (the observer and the source) move towards each other, then the frequency of the vibrations received by the observer will be higher compared to the original frequency.


Answer: 1

(a) Both A and B are correct.

Sound is a longitudinal wave produced by the oscillation of pressure at a point, thus, forming compressions and rarefactions. That portion of gas itself does not move but the pressure variation causes a disturbance.


Answer: 2

(d) p=∑p0n sin knx-ωnt.

When we clap, there is a change in pressure, which sets a disturbance and forms a wave. However, this variation is not uniform every time we clap (unlike in the case of a sound wave). Hence, we sum up all the disturbances.


Answer: 3

(d) cannot be compared with its value in water.

If B is the bulk modulus and ρ is the density, then the velocity of sound is given by:
Velocity=Bρ
If both B and ρ are greater, then we cannot compare 2B2ρ=3B3ρ=Bρ.
For proper comparison, we need numerical values.


Answer: 4

(c) Wavelength

The velocity of a sound wave varies with temperature as follows:
v∝T
As the temperature increases, the speed also increases. However, since the frequency remains the same, its wavelength changes.


Answer: 5

(d) Frequency

When a sound or light wave undergoes refraction, its frequency remains constant because there is no change in its phase.


Answer: 6

(c) the elastic property as well as the inertia property

Propagation of any wave through a medium depends on whether it is elastic and possesses inertia. A wave needs to oscillate (elastic property) for it to be propagated and if it does not have inertia, the oscillations won’t keep on moving to and fro about the mean position.


Answer: 7

(a) P1=P2

Since the average power transmitted by a wave is independent of the wavelength, we have P1=P2.


Answer: 8

(d) the energy is redistributed and the distribution remains constant in time.

The energy is redistributed due to the presence of interference. However, as the frequency and phase remain constant , the distribution also remains constant with time.


Exercise : Solution of Questions on page Number : 352


Answer: 1

Given:
Velocity of sound in air v = 330 m/s
Velocity of sound through the steel tube vs = 5200 m/s
Here, Length of the steel tube S = 1 m
As we know, t=Sv
t1=1330 and t2=15200

Where, t1 is the time taken by the sound in air.
t2 is the time taken by the sound in steel tube.
Therefore,
Required time gap t=t1-t2⇒t=1330-15200⇒t= 2.75×10-3 s⇒t= 2.75 ms

Hence, the time gap between two hearings is 2.75 ms.


Answer: 9

(b) at the middle of the pipe

For an open organ pipe in fundamental mode, an anti-node is formed at the middle, where the amplitude of the wave is maximum. Hence, the pressure variation is also maximum at the middle.


Answer: 10

(a) longitudinal stationary waves

An open organ pipe has sound waves that are longitudinal. These waves undergo repeated reflections till resonance to form standing waves.


Answer: 11

c) υ

If v is the velocity of the wave and L is the length of the pipe,
then the fundamental frequency for an open organ pipe is
ν=v2L

For a closed organ pipe of length L’ = L/2, the fundamental frequency is
ν=v4L’=v×24×L=v2L=v

(When the pipe is dipped in water, it behaves like a closed organ pipe that is half the length)


Answer: 12

(c) for both longitudinal and transverse waves

When two or more waves of slightly different frequencies (v1 – v2 ≯ 10) travel with the same speed in the same direction, they superimpose to give beats. Thus, the waves may be longitudinal or transverse.

Answer: 13

The frequency of the sonometer may be 512 ± 6Hz, i.e., 506 Hz or 518 Hz.

On increasing the tension in a sonometer wire, the velocity of the wave (v) increases proportionately as the number of beats decreases. Therefore, the frequency of the sonometer wire is 506 Hz.


Answer: 14

(d) =v

For the Doppler effect to occur, there must be relative motion between the source and the observer. However, this is not the case here. Hence, the frequency heard by the passenger is υ.


Answer: 15

(d) separation between the source and the observer

v0=v±u0v±usvs
It is clear from the equation that the change in frequency due to Doppler effect depends only on the relative motion and not on the distance between the source and the observer.


Answer: 16

At B, the velocity of the source is along the line joining the source and the observer. Therefore, at B, the source is approaching with the highest velocity as compared to A and C. Hence, the frequency heard is maximum when the source is at B.


Answer: 1

d) Wave velocity

The frequency, wavelength and amplitude do not have a unique value in the sound produced.
The frequency (and wavelength) changes as the pitch of the sound varies, while the amplitude is different as the loudness varies. However, the speed of sound in the air at a particular temperature is constant, i.e., it has a unique value.


Answer: 2

(a) larger wavelength
(c) larger velocity

The velocity varies with temperature as v∝T. Therefore, it increases.
Since the frequency remains constant, the wavelength will increase as λ∝v.


Answer: 3

(b) The first overtone may be 400 Hz.
(c) The first overtone may be 600 Hz.
(d) 600 Hz is an overtone.

For an open organ pipe:
νn=nν1

nth harmonic = (n – 1)th overtone

ν1=200 Hz, ν2=400 Hz, ν3=600 Hz

If the pipe is an open organ pipe, then the 1st overtone is 400 Hz. Option (b) is correct.

Also, υ3 = 600 Hz, i.e., second overtone = 600 Hz.
600 Hz is an overtone. Therefore, option (d) is correct.

If the pipe is a closed organ pipe, then νn=2n-1ν1.

(2n – 1)th harmonic = (n – 1)th overtone

For n = 2:
1st overtone = 3rd harmonic = 3υ1
=3 × 200
= 600 Hz
Therefore, option (c) is also correct.


Answer: 4

(c) The wavelength of the sound in the medium towards the observer decreases.

Due to Doppler effect, the frequency or wavelength of the sound changes towards the observer only.
The actual frequency and wavelength of the source does not change.


Answer: 5

(a) Frequency
(d) Time period

The frequency does not change. Hence, the time period (inverse of frequency) also remains the same.
Due to wind, the relative velocity of sound changes. Thus, the wavelength also changes so as to keep the frequency the same. (As v=νλ)


Exercise : Solution of Questions on page Number : 353


Answer: 2

Given:
The distance of the building from the meeting is 80 m.
Velocity of sound in air v = 320 ms−1
Total distance travelled by the sound after echo is S = 80 × 2 = 160 m
As we know, v=St.
∴ t=sv=160320=0.5 s
Therefore, the maximum time interval will be 0.5 seconds.


Answer: 3

Given:
Distance of the large wall from the man S = 50 m
​He has to clap 10 times in 3 seconds.
So, time interval between two claps will be
=310 second.

Therefore, the time taken t by sound to go to the wall is
t=320 second.

We know that:Velocity v=St

⇒ v=50320=333 m/s

Hence, the velocity of sound in air is 333 m/s.


Answer: 4

Given:
Speed of sound v = 360 ms−1

(a) We know that frequency∝1Wavelength.

Therefore, for minimum wavelength, the frequency f = 20 kHz.

We know that v = fλ.

∴ λ=36020×103⇒λ=18×10-3 m=18 mm

(b) For maximum wave length:

Frequency f=20 Hz

v=fλ∴λ=vf ⇒λ=36020=18 m


Answer: 5

Given:
Speed of sound in water v = 1450 ms−1
Audible range for average human ear = (20-20000 Hz)
Relation between frequency (f) and wavelength (λ) with constant velocity:
f∝1λ
(a) For minimum wavelength, the frequency should be maximum.

Frequency f = 20 kHz

As v=fλ,
∴ λ=vf.⇒ λ=145020×103⇒λ=7.25 cm

(b) For maximum wave length, the frequency should be minimum.

f = 20 Hz
v=fλ⇒ λ=145020=72.5 m
∴ λ = 72.5 m


Answer: 6

Given:
The diameter of the loudspeaker is 20 cm.
Velocity of sound in air v = 340 m/s
As per the question,
wavelength (λ) of the sound is 10 times the diameter of the loudspeaker.
∴ (λ) = 20 cm×10 = 200 cm = 2 m

(a) Frequency f = ?

As we know, v=fλ.

∴ f=vλ=3402=170 Hz

(b) Here, wavelength is one tenth of the diameter of the loudspeaker.
⇒ λ = 2 cm = 2 × 10−2 m

∴  f=vλ=3402×10-2=17,000 Hz = 17 kHz


Answer: 7

(a) Given:
Frequency of ultrasonic wave f = 4.5 MHz = 4.5 × 106 Hz
Velocity of air v = 340 m/s
Speed of sound in tissue = 1.5 km/s
Wavelength λ = ?
As we know, v=fλ.
∴ λ=3404.5×106⇒λ=7.6×10-5 m

(b) Velocity of sound in tissue vtissue= 1500 m/s
λ=vtissuef⇒ λ=15004.5×10-6 m⇒ λ=3.3×10-4 m


Answer: 8

Given:
Equation of a travelling sound wave is y = 6.0 sin (600 t − 1.8 x),
where y is measured in 10−5 m,
t in second,
x in metre.
Comparing the given equation with the wave equation, we find:
Amplitude A = 6×10-5 m

(a) We have: 2πλ=1.8 ⇒λ=2π1.8 So, required ratio : Aλ=6.0×(1.8)×10-5m/s(2π)=1.7×10-5 m

(b) Let Vy be the velocity amplitude of the wave.
Velocity v=dydtv=d6 sin 600 t-1.8 xdt⇒v=3600 cos (600t-1.8x)×10-5 m/s Amplitute Vy=3600×10-5m/s Wavelength : λ=2π1.8 Time period : T=2πω⇒ T=2π600 Wave speed v=λT⇒v=6001.8=1003 m/s Required ratio : Vyv=3600×3×10-51000=1.1×10-4 m


Answer: 9

Given:
Speed of sound in air v = 350 m/s
Frequency of sound wave f = 100 Hz
a) As we know, v=fλ.
∴ λ=vf⇒λ=350100=3.5 m
Distance travelled by the particle:
Δx = (350 × 2.5 × 10−3) m

Phase difference is given by:
ϕ=2πλ×∆xOn substituting the values we get : ϕ=2π×350×2.5×10-33.5⇒ ϕ=π2
(b) For the second case:
Distance between the two points:
∆x = 10 cm = 0.1 m
⇒ ϕ=2πλ∆x On substituting the respective values in the above equation, we get : ϕ=2π×0.13.5=2π35
The phase difference between the two points is 2π35.


Answer: 10

Given:
Separation between the two point sources ∆x = 10 cm
Wavelength λ = 5.0 cm

(a)
Phase difference is given by : ϕ=2πλ∆xSo,ϕ=2π5×10=4π

Therefore, the phase difference is zero.

(b) Zero : the particles are in the same phase since they have the same path.


Answer: 11

Given:
Pressure of oxygen p = 1.0 × 105 Nm−2
Temperature T = 273 K
Mass of oxygen M = 32 g
Volume of oxygen V = 22.4 litre = 22.4×10-3 m3
Molar heat capacity of oxygen at constant volume Cv = 2.5 R
Molar heat capacity of oxygen at constant pressure Cp = 3.5 R
Density of oxygen ρ=MV=32 g22.4×10-3 m3
We know that:CpCv=γ
∴ γ=3.5 R2.5 R=1.4 Velocity of sound is given by :  v=γpρ, where v is the speed of sound.On substituting the respective values in the above formula,
we get : v=1.4×1.0×1053222.4⇒v=310 m/s

Therefore, the speed of sound in oxygen is 310 m/s.


Answer: 12

Given:
Velocity of sound v1 = 340 m/s
Temperature T1 = 17°C = 17 + 273 = 290 K
Let the velocity of sound at a temperature T2 be v2.
T2 = 32°C = 273 + 32 = 305 K
Relation between velocity and temperature:
v∝TSo,v1v2=T1T2⇒v2=v1×T2T1 On substituting the respective values, we get : v2=340×305290=349 m/s
Hence, the final velocity of sound is 349 m/s.


Answer: 13

Let the speed of sound T1 be v1,
where T1 = 0˚ C = 273 K.
Let T2 be the temperature at which the speed of sound (v2) will be double its value at 0˚ C.
As per the question,
v2 = 2v1.
v∝T
∴​ v22v12=T2T1⇒2v12v12=T2273⇒T2=273×4 =1092 K
To convert Kelvin into degree celsius:
T2=273×4 -273 =819° C
Hence, the temperature (T2 ) will be 819˚ C


Answer: 14

Given:
The absolute temperature of air in a region increases linearly from T1 to T2 in a space of width d.
The speed of sound at 273 K is v.
vT is the velocity of the sound at temperature T.
Let us find the temperature variation at a distance x in the region.
Temperature variation is given by:
T=T1+T2-T1dx v∝T
⇒vTv=T273
⇒ vT=vT273⇒dt=dxvT=duv×273T
⇒ t=273v∫0ddxT1+T2-T1dx12
⇒t=273v×2dT2-T1T1+T2-T1d0d
⇒t=273v×2dT2-T1T2-T1
⇒t=2dv273T2-T1×T2-T1
∵ A2-B2=A-BA+B
⇒T=2dv273T2+T1 …(i)

Evaluating this time:
Initial temperature T1 = 280 K
Final temperature T2 = 310 K
Space width d = 33 m
v = 330 m s−1

On substituting the respective values in the above equation, we get:
T=2×33330273280+310=96 ms


Answer: 15

Given:
Volume of kerosene V = 1 litre = 1×10-3 m3
Pressure applied P = 2.0 × 105 Nm-2
Density of kerosene ρ = 800 kgm−3
Speed of sound in kerosene v = 1330 ms−1
Change in volume of kerosene ∆V = ?
The velocity in terms of the bulk modulus K and density ρ is given by:
v=Kp,
where K=v2ρ.
⇒K=13302×800 N/m2As we know, K=FA∆VV.
∴ ∆V=Pressure×VK
∵ P= FA On substituting the respective values, we get :
∆V=2×105×1×10-31330×1330×800=0.14 cm3

Therefore, the change in the volume of kerosene ∆V = 0.14 cm3.


Answer: 16

Given:
Wavelength of sound wave λ = 35 cm = 35×10-2 m
Pressure amplitude P0 = 1.0×105±14 Pa
Displacement amplitude of the air particles S0 = 5.5 × 10−6 m
Bulk modulus is given by:
B=P0λ2πS0=∆p∆V/V
On substituting the respective values in the above equation, we get:
B=14×35×10-2 m2π5.5×10-6 m⇒B=1.4×105 N/m2

Hence, the bulk modulus of air is 1.4×105 N/m2.


Answer: 17

Given:
Velocity of sound in air v = 340 ms−1
Power of the source P = 20 W
Frequency of the source f = 2,000 Hz
Density of air ρ = 1.2 kgm −3

(a) Distance of the source r = 6.0 m
Intensity is given by:
I=PA,
where A is the area.
⇒I=204πr2=204×π×62 ∵r=6 m⇒I=44 mw/m2

(b) As we know,
I=p022ρv.⇒ P0=I×2ρv⇒P0=2×1.2×340×44×10-3⇒P0=6.0 Pa or N/m2
(c) As we know, I = 2π2S02v2ρV.
S0 is the displacement amplitude.
⇒ S0=I2π2v2ρV
On applying the respective values, we get:
S0 = 1.2 × 10−6 m


Answer: 18

Given:
The intensity I1 is 1.0 × 10−8 Wm−2,
when the distance of the point source r1 is 5 m.
Let I2 be the intensity of the point source at a distance r2 = 25 m.
As we know,
I ∝1r2.So,I1I2=r22r11⇒ I2=I1r12r22
On substituting the respective values, we get:

I2=1.0×10-8×25625 =4.0×10-10W/m2


Answer: 19

Let βA be the sound level at a point 5 m (= r1) away from the point source and βB be the sound level at a distance of 50 m (= r2) away from the point source.
∴​ βA = 40 dB
Sound level is given by:
β=10log10II0
According to the question,
βA=10 log10 IAI0.⇒ IAI0=10βA10 …..1
βB=10 log10IBIo⇒ IBI0=10βB10 …..2
From 1 and 2, we get: IAIB=10βA-βB10 ….3
Also, IAIB=rB2rA2=5052 = 102 …..4
From 3 and 4,
we get : 102=10βA-βB10⇒ βA-βB10=2 ⇒ βA-βB=20⇒ βB=40-20=20 dB

Thus, the sound level of a point 50 m away from the point source is 20 dB.


Answer: 20

Let the intensity of the sound be I and β1 be the sound level. If the intensity of the sound is doubled, then its sound level becomes 2I.
Sound level β1 is given by:
β1=10 log10II0,
where I0 is the constant reference intensity.
When the intensity doubles, the sound level is given by:
β2=10 log102II0.
According to the question,
β2-β1=10 log2II =10×0.3010=3 dB

The sound level is increased by 3 dB.


Answer: 21

Given:
The sound level that can hurt the human ear is 120 dB. Then, the intensity I is 1 W/m2.
Audio output of the small speaker P = 2 W
Let the closest distance be x.
We have:
I=P4πr224πx2=1
⇒ x2=24π⇒ x=0.4 m=40 cm

Hence, the closest distance of the human ear from the small speaker is 40 cm.


Answer: 22

Given:
Initial sound level β1 = 50 dB
Final sound level β2 = 60 dB
Constant reference intensity I0 = 10-12 W/m2
We can find initial intensity I1 using:
β1=10log10I1I0.
⇒50 =10log10I110-12
On solving, we get:
I1 = 10-7 W/m2.
Similarly,
β2=10log10I2I0.
On substituting the values and solving, we get:
I2=10-6 W/m2
As the intensity is proportional to the square of pressure amplitude (p),
we have:
I2I1=p2p12=10-610-7=10
∴ p2p12=10⇒ p2p1=10
Hence, the pressure amplitude is increased by 10 factor.


Answer: 23

Let the intensity of each student be I and the sound level of 50 students be β1. If the number of students increases to 100, the sound level becomes β2.
Using β=10log10II0,
where I0 is the constant reference intensity, I is the intensity and β is the sound level.
β1=10log1050II0β2=10log10100II0
⇒β2-β1=10log10100II0-10log1050II0 =10log10100I50I =10log102 =3

Therefore, the noise level of 100 students β2 will be = 50+3 = 53 dB.


Answer: 24

Given:
Speed of sound in air v = 340 ms−1
Distance moved by sliding tube = 2.50 cm
Frequency of sound f = ?

Distance between maximum and minimum: λ4=2.50 cm⇒ λ=2.50×4=10 cm=10-1m

As we know,
v = fλ.

∴  f=vλ⇒f=34010-1= 3400 Hz=3.4 kHz

Therefore, the frequency of the sound is 3.4 kHz.


Answer: 25

The sliding tube is pulled out by a distance of 16.5 mm.
Speed of sound in air, v = 330 ms−1

(a) As per the question, we have:
λ4=16.5 mm⇒ λ=16.5×4=66 mm=66×10-3 m

We know:
v = f λ
∴  f=vλ
⇒ f=vλ=34066×10-3=5 kHz

(b)​
Ratio of maximum intensity to minimum intensity:
IMaxIMin=KA1-A22KA1+A22=I9I⇒A1-A22A1+A22=19Taking square roots of both sides, we get:
A1+A2A1-A2=31⇒A1A2=3+13-1=21
So, the ratio of the amplitudes is 2.


Exercise : Solution of Questions on page Number : 354


Answer: 26

Given:
Speed of sound in air v = 320 ms−1
The path difference of the sound waves coming from the loudspeaker and reaching the person is given by:
Δx = 6.4 m − 6.0 m = 0.4 m
If f is the frequency of either wave, then the wavelength of either wave will be:
λ=vf=320f
For destructive interference, the path difference of the two sound waves reaching the listener should be an odd integral multiple of half of the wavelength.
∴ ∆x=(2n+1)λ2 , where n is an integer.
On substituting the respective values, we get:
0.4 m=2n+1×3202f⇒f=2n+13202×0.4⇒f=(2n+1) 400 Hz

Thus, on applying the different values of n, we find that the frequencies within the specified range that caused destructive interference are 1200 Hz, 2000 Hz, 2800 Hz, 3600 Hz and 4400 Hz.


Answer: 27

Given:
Velocity of sound in air v = 336 ms−1
Distance between maximum and minimum intensity: λ4 = 20 cm
Frequency of sound f = ?

We have:
λ4=20⇒λ=20×4=80 cm=80×10-2 m

As we know, v=fλ.
∴  f=vλ
⇒ f=33680×10-2=420 Hz

Therefore, the frequency of the sound emitted from the source is 420 Hz.


Answer: 28

Given:
Distance between the source and detector = d
Distance of cardboard from the source = 2d
Wavelength of the source λ = d/2
Path difference between sound waves received by the detector before shifting the cardboard:
2d22+2d2-d⇒2×3d2-d⇒2d
If the cardboard is shifted by a distance x, the path difference will be:

2d22+2d+x2-d
According to the question,

2d22+2d+x2-d=2d+d4⇒2d22+2d+x2-d=9d4⇒2d22+2d+x2=9d4+d=13d4⇒ d22+2d+x2=16964d2⇒ 2d+x2=(169-16)64d2=15364d2⇒2d+x=1.54d⇒x=1.54-1.41d=0.13d


Answer: 29

Given:
Distance between the two speakers d = 2.40 m
Speed of sound in air v = 320 ms−1
Frequency of the two stereo speakers f = ?

As shown in the figure, the path difference between the sound waves reaching the listener is given by:
∆x=S2L-S1L
∆x=(3.2)2+(2.4)2-3.2

Wavelength of either sound wave:
=320f
We know that destructive interference will occur if the path difference is an odd integral multiple of the wavelength.
∴ ∆x=(2n+1)λ2
So,
(3.2)2+(2.4)2-3.2=(2n+1)2320f⇒16-3.2=2n+12320f⇒0.8×2f=2n+1×320⇒ 1.6f=2n+1×320⇒f= 200(2n+1)

On putting the value of n = 1,2,3,…49, the person can hear in the audible region from 20 Hz to 2000 Hz.


Answer: 30

Given:
Wavelength of sound wave λ = 20 cm
Separation between the two sources AC = 20 cm
Distance of detector from source BD = 20 cm

If the detector is moved through a distance x, then the path difference of the sound waves from sources A and C reaching B is given by:

Path difference = AB – BC
= 202+10+x2 – 202+10-x2

To hear the minimum, this path difference should be equal to:
2n+1λ2 =λ2 = 10 cm
So,
202+10+x2-202+10-x2 = 10

On solving, we get, x = 12.6 cm.

Hence, the detector should be shifted by a distance of 12.6 cm.


Answer: 31

Given:
Frequency of source f = 600 Hz
Speed of sound in air v = 330 m/s
v=fλ
∴  λ=vf=330600=0.5 mm
Let the man travel a distance of y parallel to the y-axis and let d be the distance between the two speakers. The man is standing at a distance of D from the origin.
The path difference (x) between the two sound waves reaching the man is given by:
x=S2Q-S1Q=ydD
Angle made by man with the origin:
θ=yD
Given:
d = 2

The destructive interference of sound (minimum intensity) takes place if the path difference is an odd integral
multiple of half of the wavelength.

 

(b) For maximum intensity:
The constructive interference of sound (maximum intensity) takes place if the path difference is an integral multiple of the wavelength.
x=nλFor n=1 : ⇒ydD=λ⇒θ=λd⇒θ=0.552=0.275 rad∴ θ=16°
(c) The more number of maxima is given by the path difference:
ydD=2λ, 3λ, 4λ,…..
⇒ yD=θ=32°, 64°, 128°

He will hear two more maxima at 32° and 64° because the maximum value of θ may be 90°.


Answer: 32

Given:
All the three sources of sound, namely, S1, S2 and S3 emit equal intensity of sound waves.
Therefore, all three sources have equal amplitudes.
∴  A1 = A2 = A3
Now, by vector method, the resultant of the amplitudes is 0.
So, the resultant intensity at P is zero.

 


Answer: 33

Given:
S1& S2 are in the same phase. At O, there will be maximum intensity.
There will be maximum intensity at P.
From the figure (in questions):

∆S1PO and ∆S2PO are right-angled triangles.
So,
S1P2-S2P2=D2+x2-D-2λ2+x22=4λD+4λ2=4λD(λ2is small and can be neglected)⇒S1P+S2PS1P-S2P=4λD⇒S1P-S2P=4λDS1P+S2P⇒ S1P-S2P=4λD2x2+D2
For constructive interference, path difference = nλ.
So,
⇒ S1P-S2P=4λD2x2+D2=nλ⇒ 2Dx2+D2=n⇒ n2(x2+D2)=4D2⇒ n2x2+n2D2=4D2⇒n2x2=4D2-n2D2⇒n2x2=D24-n2⇒ x=Dn4-n2When n=1, x=3D (1st order).When n=2,x=0 (2nd order).
So, when x = 3D , the intensity at P is equal to the intensity at O.


Answer: 34

Let the sound waves from the two coherent sources S1 and S2 reach the point P.

rework
OQ = R cosθ
OP = R cosθ
OS2 = OS1 = 1.5 λ
From the figure, we find that:
PS12=PQ2+QS2=Rsinθ2+Rcosθ-1.5λ2
PS12=PQ2+QS12=Rsinθ2+Rcosθ+1.5λ2
Path difference between the sound waves reaching point P is given by:
S1P2-S2P2=Rsinθ2 + Rcosθ +1.5λ2-Rsinθ2+Rcosθ-1.5λ2 =1.5λ+Rcosθ2-Rcosθ-1.5λ2 =6λRcosθ

S1P-S2P=6λR cosθ2R = 3λ cosθ
Suppose, for constructive interference, this path difference be made equal to the integral multiple of λ.
Hence,
S1P-S2P=3λ cosθ=nλ⇒ cosθ=n3⇒ θ= cos-1n3
where, n = 0, 1, 2, …

⇒ θ = 0°, 48.2°, 70.5°and 90° are similar points in other quadrants.


Exercise : Solution of Questions on page Number : 354


Answer: 35

Given:
Resultant intensity at P = I0
The two sources of sound S1 and S2 vibrate with the same frequency and are in the same phase.
(a) When θ = 45°:
Path difference = S1P − S2P = 0

So, when the source is switched off, the intensity of sound at P is I04.
(b) When θ = 60°, the path difference is also 0. Similarly, it can be proved that the intensity at P is I04 when the source is switched off.


Answer: 36

Given:
Speed of sound in air v = 340 m/s
Length of open organ pipe L = 20 cm = 20 × 10−2 m
Fundamental frequency f of an open organ pipe:
f=v2L=3402×20×10-2=850 Hz

First overtone frequency f1:
f1 = 2f
⇒f1=2V2I=2×850=1700 Hz

Second overtone frequency f2:
f2=3f
⇒f2=3V2L=3×850=2550 Hz


Answer: 37

Given:
Speed of sound in air v = 340 ms−1
Frequency of closed organ pipe f = 500 Hz
Length of tube L = ?
Fundamental frequency of closed organ pipe is given by:
f=v4L
∴  L=v4f∴ L=3404×500=0.17 m=17 cm


Answer: 38

Given:
Distance between two nodes = 4 cm
Speed of sound in air v = 328 ms−1
Frequency of source f = ?
Wavelength λ = 2 × 4.0 = 8 cm
v = fλ
∴  f=vλ=3288×10-2=4.1 KHz .

Hence, the required frequency of the source is 4.1 KHz.


Answer: 39

Given:
Separation between the node and anti-node = 25 cm
Speed of sound in air v = 340 ms−1
Frequency of vibration of the air column f = ?
The distance between two nodes or anti-nodes is λ.
We have:
λ4=25 cm⇒ λ=100 cm=1 m
Also,
v=fλ
⇒ f=vλ=3401=340 Hz
Hence, the frequency of vibration of the air column is 340 Hz.


Answer: 40

Given:
Length of cylindrical metal tube L = 50 cm
Speed of sound in air v = 340 ms−1
Fundamental frequency f1 of an open organ pipe:
f1=v2L=3402×50×10-2=340 Hertz

So, the required harmonics will be in the range of 1000 Hz to 2000 Hz.

f2=2×340=680 Hzf3=3×340=1020 Hzf4=4×340=1360 Hzf5=5×340=1700 Hzf6=6×340=2040 Hz

f2, f3, f4… are the second, third, fourth overtone, and so on.
The possible frequencies between 1000 Hz and 2000 Hz are 1020 Hz, 1360 Hz and 1700 Hz.


Answer: 41

Given:
Length of air column at first resonance L1 = 20 cm = 0.2 m
Length of air column at second resonance L2 = 62 cm = 0.62 m
Frequency of tuning fork f = 400 Hz

(a) We know that:
λ=2L2-L1⇒ λ=262-20=84 cm=0.84 m

v = λf,
where v is the speed of the sound in air.
So,
v=0.84×400=336 m/s
Therefore, the speed of the sound in air is 336 m/s.

(b) Distance of open node is d:

L1+d=λ4⇒d=λ4-L1=21-20=1 cm
Therefore, the required distance is 1 cm.


Answer: 42

Given:
Length of closed organ pipe L1 = 30 cm
Length of open organ pipe L2 = ?
Let f1 and f2 be the frequencies of the closed and open organ pipes, respectively.
The first overtone frequency of a closed organ pipe P1 is given by
f1=3v4L1,
where v is the speed of sound in air.
On substituting the respective values, we get:
f1=3v4×30
Fundamental frequency of an open organ pipe is given by:
f2=v2L2
As per the question,
f1=f2 3×v4×30=v2L2 ⇒ L2=20 cm
∴ The length of the pipe P2 will be 20 cm.


Answer: 43

Given:
Length of copper rod l = 1.0 m
Speed of sound in copper v = 3.8 kms−1 = 3800 m/s
Let f be the frequency of the longitudinal waves.

Wavelength λ will be:
λ2=I⇒ λ=2I=2×1=2 m

We know that:
v = fλ
⇒f=vλ
So,
f=38002=1.9 KHz
Therefore, the frequencies between 20 Hz and 20 kHz that will be heard are
= n × 1.9 kHz,
where n = 0, 1, 2, 3, …10.


Answer: 44

Given:
Speed of sound in air v = 340 ms−1
We are considering a minimum fundamental frequency of f = 20 Hz,
since, for maximum wavelength, the frequency is a minimum.
Length of organ pipe l = ?
We have:
λ2=I⇒ λ=2I

We know that:
v = fλ
⇒ λ=vf⇒l=v2×f

On substituting the respective values in the above equation, we get:

I=3402×20⇒ l=344=8.5 m
Length of the organ pipe is 8.5 m.


Answer: 45

Given:
Length of organ pipe L = 5 cm = 5 × 10−2 m
v = 340 m/s
The audible range is from 20 Hz to 20,000 Hz.
The fundamental frequency of an open organ pipe is:
f=v2L
On substituting the respective values ,we get:
f=3402×2×10-2=3.4 kHz

(b) If the fundamental frequency is 3.4 kHz, then the highest harmonic in the audible range (20 Hz – 20 kHz) is

Required highest harmonic = 20,0003400=5.8=5


Answer: 46

Given:
Length of air column in the tube l = 80 cm = 80 × 10−2 m
Speed of sound in air v = 320 ms−1
The frequency of the loudspeaker can be varied between 20 Hz to 2 KHz.
The resonance column apparatus is equivalent to a closed organ pipe.
Fundamental note of a closed organ pipe is given by:
f=v4l
⇒ f=3204×50×10-2=100 Hz

So, the frequency of the other harmonics will be odd multiples of f = (2n + 1)100 Hz.
According to the question, the harmonic should be between 20 Hz and 2 kHz.
∴ n = (0, 1, 2, 3, 4, 5, ….. 9)


Answer: 47

Given:
Velocity of sound in air v = 324 ms−1
Let l be the length of the resonating column.
Then, the frequencies of the two successive resonances will be (n+2)v4I and nv4I .
As per the question,

n+2v4l = 2592

nv4l = 1944

So,

(n+2)v4l-nv4I=2592-1944=648⇒2v4l=648⇒l=2×324×1004×648cm=25 cm

Hence, the length of the tube is 25 cm.


Answer: 48

Given:
Frequency of tuning fork f = 512 Hz
Let the speed of sound in the tube be v.
Let l1 be the length at which the piston resonates for the first time and l2 be the length at which the piston resonates for the second time.
We have:
l2 =2l1 = 2×32 = 64 cm =0.64 m
Velocity v = f×l2
⇒ v = 512 × 0.64 = 328 m/s

Hence, the speed of the sound in the tube is 328 m/s.


Answer: 49

Given:
Speed of sound in air v = 330 ms−1
Frequency of the tuning fork f = 440 Hz

For the shorter arm:
Let the length of the shorter arm of the tube be L1 .
Frequency of fundamental mode is given by:
f=v4L1

On substituting the respective values, we get:
440=3304L1 ⇒L1=330440×4=0.1875 m=18.8 cm

For the longer arm:
Let the length of the longer arm of the tube be L2 .
Frequency of the first overtone f = 440 Hz
Frequency of the first overtone is given by:
f=3v4L2

On substituting the respective values, we get:

440=3×3304L2 ⇒L2=3×330440×4=0.563 m =56.3 cm


Answer: 50

Given:
Speed of sound in air v = 340 ms−1
Length of the wire l = 40 cm = 0.4 m
Mass of the wire M = 4 g

Mass per unit length of wire m is given by:

m=MassUnit length=10-2 kg/m

n0 = frequency of the tuning fork
T = tension of the string

Fundamental frequency:
n0=12LTm

For second harmonic, n1=2n0:

n1=22LTm …..i

n1= 2n0=3404×1=85 Hz
On substituting the respective values in equation (i), we get:

85=22×0.4T10-2⇒T=(85)2×(0.4)2×10-2 =11.6 Newton

Hence, the tension in the wire is 11.6 N.


Answer: 51

Given:
Mass of long wire M = 10 gm = 10 × 10−3
Length of wire l = 30 cm = 0.3 m
Speed of sound in air v = 340 m s−1

Mass per unit length m is
m=MassUnit length=33×10-3kg/m

Let the tension in the string be T.

The fundamental frequency n0 for the closed pipe is
n0=v4I=3402×30×10-2=170 Hz

The fundamental frequency n0 is given by:
n0=12lTm
On substituting the respective values in the above equation, we get:
170=12×30×10-2×T33×10-3⇒ T=347 Newton

Hence, the tension in the wire is 347 N.


Answer: 52

Let f be the frequency of an open pipe at a temperature T. When the fundamental frequency of an organ pipe changes from v to v + ∆v, the temperature changes from T to T + ∆T.

We know that:
ν∝ T …..i

According to the question,

ν+∆ν∝ ∆T+T.

Applying this in equation (i), we get:

ν+∆νν=∆T+TT
1+∆νν=1+∆TT1/2

By expanding the right-hand side of the above equation using the binomial theorem, we get:

1+∆νν=1+12×∆TT (neglecting the higher terms)

∆νν=12∆TT


Exercise : Solution of Questions on page Number : 355


Answer: 53

Given:
The fundamental frequency of a closed pipe is 293 Hz. Let this be represented by f1.
Temperature of the air in closed pipe T1 = 20°C = 20 + 273 = 293 K
Let f2 be the frequency in the closed pipe when the temperature of the air is T2 .
∴​ T2 = 22°C = 22 + 273 = 295 K

Relation between f and T:

f∝T
∴ f1f2=T1T2⇒ 293f2=293295⇒ f2=293×295293=294 Hz


Answer: 54

Given:
Speed of sound in air vair = 340 ms−1
Velocity of sound in Kundt’s tube vrod = ?
Length at which copper rod is clamped l = 25 cm = 25×10-2 m
Distance between the heaps ∆l = 5 cm = 5×10-2 m

vrodvair=2l∆l⇒vrod = 2l∆l×vairOn substituting the respective values in the above equation, we get: vrod=340×25×10-2×25×10-2⇒ vrod=3400 m/s


Answer: 55

Given:
Length at which steel rod is clamped l = 12=0.5 m
Fundamental mode of frequency f = 2600 Hz
Distance between the two heaps ∆l = 6.5 cm = 6.5×10-2 m

Since Kundt’s tube apparatus is a closed organ pipe, its fundamental frequency is given by:

f=vair4L⇒vair=f×2×∆L⇒vair=2600×2×6.5×10-2=338 m/s(b) vsteelvair=2×l∆l⇒vsteel =2l∆l×vair⇒ vsteel=2×0.5×3386.5×10-2⇒ vsteel=5200 m/s

 


Answer: 56

Given:
First Frequency f1 = 476 Hz
Second frequency f2 = 480 Hz
Number of beats produced per second by the tuning fork m = 2

As the tuning fork produces 2 beats, its frequency should be an average of two.
This is given by :

f=f1+f22f=476+4802=478 Hz


Answer: 57

Frequency of tuning fork A:
n1 = 256 Hz

No. of beats/second m = 4

Frequency of second fork B: n2 =?
n2=n1±m
⇒n2=256±4
⇒n2 = 260 Hz or 252 Hz

Now, as it is loaded with wax, its frequency will decrease.
As it produces 6 beats per second, the original frequency must be 252 Hz.
260 Hz is not possible because on decreasing the frequency, the beats per second should decrease, which is not possible.


Answer: 58

For source A :
Wavelength λ = 32 cm = 32×10-2 m
Velocity v = 350 ms-1

Frequency n1 is given by:
n1=vλ=35032×10-2=1093.75 Hz

For source B :
Velocity v = 350 ms−1
Wavelength λ = 32.2 cm = 32.2×10-2 m

Frequency n2 is given by :
n2=vλ=35032.2×10-2=1086.96 Hz

∴ Beat frequency = 1093.75 − 1086.96 = 6.79 Hz ≈7 Hz


Answer: 59

Given:
Length of the closed organ pipe L = 40 cm = 40 × 10−2 m
Velocity of sound in air v = 320 ms−1
Frequency of the fundamental note of a closed organ pipe n is given by:

n=v4L

⇒​ n=v4L=3204×40×10-2=200 Hz

As the tuning fork produces 5 beats with the closed pipe, its frequency must be 195 Hz or 205 Hz.
The frequency of the tuning fork decreases as and when it is loaded. Therefore, the frequency of the tuning fork should be 205 Hz.


Answer: 60

Mass per unit length of both the wires​ = m

Fundamental frequency of wire of length l and tension T is given by:

n=12ITm

It is clear from the above relation that as the tension increases, the frequency increases.

Fundamental frequency of wire A is given by:Answer: 62

nA=12ITAm

Fundamental frequency of wire B is given by:

nB=12ITBm

It is given that 6 beats are produced when the tension in A is increased.

⇒​ nA=606=12lTAm
Therefore, the ratio can be obtained as:

nAnB=606600=1/2ITA/m1/2ITB/m⇒606600=TATB⇒ TATB=606600=1.01⇒ TATB=1.02


Answer: 61

Given:
Length of the wire l = 25 cm = 25 × 10−2 m
Frequency of tuning fork f = 256 Hz
Let T be the tension and m the mass per unit length of the wire.

Frequency of the fundamental note in the wire is given by:

f=12lTm

It is clear from the above relation that by shortening the length of the wire, the frequency of the vibrations increases.
In the first case:
256=12×25×10-2Tm …(1)

Let the length of the wire be l1, after it is slightly shortened.

As the vibrating wire produces 4 beats with 256 Hz, its frequency must be 252 Hz or 260 Hz. Again, its frequency must be 252 Hz, as the beat frequency decreases on shortening the wire.

In the second case:
⇒ 252=12×I1Tm …(2)

Dividing (2) by (1), we have:

252256=I125×10-2⇒ I1=252×25×10-2256 = 0.24609 m

So, it must be shortened by (25 − 24.61)
= 0.39 cm.


Answer: 62

Velocity of sound in air v = 340 ms−1
Velocity of scooter-driver vo = 36 kmh−1 = 36×518 = 10 ms-1
Frequency of sound of whistle fo = 2 kHz

Apparent frequency f heard by the scooter-driver approaching the policeman is given by:

f=v+vov×fo

f=340+10340×2 =350×2340 kHz = 2.06 kHz


Answer: 63

Given:
Frequency of sound emitted by horn f0 = 2400 Hz
Speed of sound in air v = 340 ms−1
Velocity of car vs = 18 kmh−1 = 18×518 m/s = 5 m/s

Apparent frequency of sound f is given by:

f=vv-vs×f0

On substituting the values, we get:

f=340340-5×2400 =2436 Hz


Answer: 64

Given:
Frequency of whistle f0=1250 Hz
Velocity of car vs = 72 kmh−1 = 72×518=20 ms-1
Speed of sound in air v = 340 ms−1

(a) When the car is approaching the person:

Frequency of sound heard by the person f1 is given by:
f1=vv-vs×f0

On substituting the given values in the above equation, we have:

f1=340340-20×1250 =1328 Hz

(b) When the person is behind the car:

Frequency of sound heard by the person f2 is given by:

f2=vv+vs×f0

On substituting the given values in the above equation, we have:

f2=340340+20×1250 =340360×1250=1181 Hz


Answer: 65

Given:
Speed of sound in air v= 332 ms−1
Velocity of train vs = 54 kmh−1 = 54×518 m/s=15 m/s

Let f0 be the original frequency of the train.

When the train approaches a platform, the frequency of sound heard by the observer f is given by:

f=vv-vs×f0

On substituting the values, we have:

1620=332332-15f0⇒ f0=1620×317332 Hz.

When the train crosses the platform, the frequency of sound heard by the observer f1 is given by:

f1=vv+vs×f0

Substituting the respective values in the above formula, we have :

f1=332332+15×1620×317332 =317347×1620=1480 Hz


Answer: 66

Given:
Speed of bat v = 6 ms−1
Frequency of ultrasonic wave f = 4.5 × 104 Hz
Velocity of bird vs = 6 ms−1
Let us assume that the bat is flying between the walls X and Y.
Apparent frequency received by the wall Y is

f’ = vv-vs×f0⇒f’=330330-6×4.5×104⇒f’=4.58×104 Hz

Now, the apparent frequency received by the bat after reflection from the wall Y is given by:

f”=v+vsv×fx⇒f”=330+6330×4.58×104⇒f”=4.66×104 Hz

Frequency of ultrasonic wave received by wall X:

n’=330330+6×4.5×104 =4.41×104 Hz

The frequency of the ultrasonic wave received by the bat after reflection from the wall X is

n”=v-vsv×n’ =330-6330×4.41×104 =4.33×104 Hz

Beat frequency heard by the bat is

=4.66×104-4.33×104=3300 Hz.


 

Answer: 67

Given:
Velocity of bullet vs = 220 ms−1
Speed of sound in air v = 330 ms−1
Let the frequency of the bullet be f.

Apparent frequency heard by the person f1 before crossing the bullet is given by:

f1 = vv-vs×f

On substituting the values, we get:
f1=330330-220×f=3f ….1

Apparent frequency heard by the person f2 after crossing the bullet is given by:

f2=vv+vs×f

On substituting the values, we get:
f2=330330+220×f=0.6f …..2
So,
f2f1=0.6f3f=0.2

∴  Fractional change = 1 − 0.2 = 0.8


Answer: 68

Given:
Speed of sound in air v = 340 ms−1
Frequency of whistles f0 = 500 Hz
Speed of train vs = 72 km/h = 72×518=20 m/s

The person will receive the sound in a direction that makes an angle θ with the track. The angle θ is given by:
θ=tan-10.52.4/2=22.62°

The velocity of the source will be ‘v cos θ’ when heard by the observer.

So, the apparent frequency received by the man from train A is

f1=vv-vscosθ×f0⇒f1=340340-vscos 22.62°×500⇒f1=340340-20×cos 22.62°×500⇒f1=528.70 Hz ≈529 Hz

The apparent frequency heard by the man from train B is

f2=vv+vcosθ×f0⇒f2=340340+20×cos 22.62°×500⇒f2=474.24 Hz ≈474 Hz


Answer: 69

Given:
Frequency of violins f0 = 440 Hz
Speed of sound in air v = 340 ms−1
Let the velocity of the train (sources) be vs.

(a) Beat heard by the standing man = 4
∴ frequency f1 = 440 + 4
=  444 Hz or 436 Hz
Now,
f1=340340-vs×f0

On substituting the values, we have:

444=340+0340-vs×440

⇒ 444340-vs=440×340
⇒ 340×444-440=440×vs⇒340×4=440×vs⇒vs =3.09 m/s=11 km/h

(b) The sitting man will listen to fewer than 4 beats/s.


Answer: 70

Given:
Speed of sound in air v = 332 ms−1
Velocity of the observer v0 = 3 ms-1
Velocity of the source vs = 0
Frequency of the tuning forks f0 = 256 Hz
The apparent frequency f1 heard by the man when he is running towards the tuning forks is

f1=v+v0v×f0

On substituting the values in the above equation, we get:

f1=332+3332×256=258.3 Hz

The apparent frequency f2 heard by the man when he is running away from the tuning forks is

f2=v-v0v×f0

On substituting the values in the above equation, we get:

f2=332-3332×256 =253.7 Hz.

∴  beats produced by them
= f2-f1
=258.3 − 253.7 = 4.6 Hz


Answer: 71

Given:
Frequency of tuning forks f0 = 512 Hz
Speed of sound in air v = 330 ms−1
Velocity of tuning forks vs = 5.5 ms−1
The apparent frequency f1 heard by the person from the tuning fork on the left is given by:

f1=vv-vs×f0
On substituting the values in the above equation, we get:

f1=330330-55×512 =520.68 Hz

Similarly, apparent frequency f2 heard by the person from the tuning fork on the right is given by:
f2=vv-vs×f0

On substituting the values in the above equation, we get :

f2=330330-5.5×512 =503.60 Hz

∴  beats produced
= f1-f2
=  520.68 − 503.60 = 17.5 Hz

As the difference is greater than 10 ( persistence of sound for the human ear is 1/10 of a second), the sound gets overlapped and the observer is not able to distinguish between the sounds and the beats.


Exercise : Solution of Questions on page Number : 356


Answer: 72

Given:
Speed of sound in air v = 332 ms−1
Radius of the circle r = 100π cm = 1π m
Frequency of sound of the source f0 = 500 Hz
Angular speed ω = 5 rev/s
Linear speed of the source is given by:
v=ωr
⇒ v=5×1π=5π=1.59 m/s
∴  velocity of source vs = 1.59 m/s

Let X be the position where the observer will listen at a maximum and Y be the position where he will listen at the minimum frequency.

Apparent frequency f1 at X is given by:

f1=vv-vsf0

On substituting the values in the above equation, we get :
f1=332332-1.59×500≈515 Hz

Apparent frequency f2 at Y is given by :

f2=vv+vsf0

On substituting the values in the above equation, we get :
f2=332332+1.59×500≈485 Hz


Answer: 73

Given:
Velocity of sound in air v = 350 ms−1
Velocity of source vs = 90 km/hour = 90×518 = 25 m/s
Velocity of observer v0 = 25 m/s
Frequency of whistle f0 = 500 Hz

Apparent frequency f heard by the observer in train B is given by:

f=v-v0v-vsf0

On substituting the respective values in the above equation, we get:

f=350+25350-25×500=577 Hz

The apparent frequency heard in the other train is 577 Hz.


Answer: 74

Given:
Frequency of whistle f0 = 16 × 103 Hz
Apparent frequency f = 20 × 103 Hz
(f is greater than that value)
Velocity of source vs = 0
Let v0 be the velocity of the observer.
Apparent frequency f is given by:

f=v+v0v-vsf0

On substituting the values in the above equation, we get :

20×103=330+v0330-0×16×103⇒ 330+v0=20×33016⇒ v0=20×330-16×3304 =3304m/s=297 km/h

(b) This speed is not practically attainable for ordinary cars.

 


Answer: 75

Given:
Velocity of car sounding a horn vs = 108 km/h = 108×518 m/s = 30 m/s
Velocity of front car v0 = 72 kmh−1 = 72×518 = 20 m/s
Frequency of sound emitted by horn f0 = 800 Hz
Velocity of air v = 330 ms−1
Apparent frequency of sound heard by driver in the front car (f) is given by:

f=v-v0v-vsf0

On substituting the values in the above equation, we get:

f=330-20330-30×800=826.67≃827 Hz


Answer: 76

Given:
Velocity of water v = 1500 m/s
Frequency of sound signal f0 = 2000 Hz
Velocity of first submarine vs = 36 kmh−1 = 36×518 m/s = 10 m/s
Velocity of second submarine v0 = 54 km h−1 = 54×518 m/s = 15 m/s
Frequency received by the first submarine f1 is given by:

f1 =v + v0v – vsf0

On substituting the values, we get:

f1 =1500+151500-10×(2000) = 2034 Hz

(b) Here, f0 = 2034 Hz.

Apparent frequency received by second submarine f2 is given by:

f2 =1500+101500-15×(2034) =2068 Hz


Answer: 77

Given:
Amplitude r = 17 cm = 17100 = 0.17 m
Frequency of sound emitted by source f = 800 Hz
Velocity of sound v = 340 m/s
Frequency band = f2 – f1= 8 Hz
Here, f2 and f1 correspond to the maximum and minimum apparent frequencies (Both will be at the mean position because the velocity is maximum).

Now, f1=340340+vsf and f2=340340-vsf
∴ f2-f1=8⇒ 8 =340340 – vsf-340340 + vsf⇒ 8 = 340f1340-vs-1340+vs
⇒ 8 =340×800×2vs3402 – vs2⇒ 2vs3402-vs2=8340×800⇒ 3402-vs2=68000 vs

Solving for vs, we get:
vs = 1.695 m/s

For SHM:

vs=rω⇒ ω=1.6950.17=10
∴ T=2πw=π5=0.63 sec


Answer: 78

Given:
Frequency of pulse produced by the bike f0 = 1650 Hz
Velocity of bike vb = 42 ms−1
Velocity of sound in air v = 334 ms−1
Frequency of pulse received by the second boy f = ?
Velocity of an observer v0 = 0
Velocity of source will be:
vs = vbcosθ

= 42×cos45°
= 42×12 = 4 ms-1

Frequency of pulse received by the second boy is given by:

f=vv – vsf0 =334334 – 4×1650 =1670 Hz


Answer: 79

Given:
Frequency of sound emitted by the source n0 = 660 Hz
Velocity of sound in air v = 330 ms-1
Velocity of observer v0 = 26 ms−1
Frequency of sound heard by observer n = ?

(a) At y = 140 m:
Frequency of sound heard by the listener, when the source is fixed but the listener is moving towards the source:

n=v+v0v n0

Here, v0 = v0cosθ

On substituting the values, we get :

n=v+v0cosθv n0=330+26×140364330×660=340×2=680 Hz

(b) When the observer is at y = 0, the velocity of the observer with respect to the source is zero.
Therefore, he will hear at a frequency of 660 Hz.

(c) When the observer is at y = 140 m :

n=v-v0v×n0
Here, v0 = v0cosθ

On substituting the values, we get :

n = 330-26×140364330×660n = 330-10330×660=640 Hz


Answer: 80

Given:
Velocity of sound in air v = 340 m/s
Velocity of source vs = 108 kmh-1 = 108×100060×60=30 ms-1
Frequency of the source n0 = 500 Hz

(a) Since the velocity of the passenger with respect to the train is zero, he will hear at a frequency of 500 Hz.

(b) Since the observer is moving away from the source while the source is at rest:

Velocity of observer vo = 0
Frequency of sound heard by person standing near the track is given by:

n=vv+vsn0

Substituting the values, we get:

n=340340+30×500=459 Hz

(c) When medium (wind) starts blowing towards the east:

Velocity of medium vm = 36 kmh-1 = 36×518 = 10 ms-1

The frequency heard by the passenger is unaffected (= 500 Hz).

However, frequency heard by person standing near the track is given by:

n=v+vmv+vm+vs×n0 =340+10340+10+30×500 =458 Hz


Answer: 81

Given:
Velocity of sound in air v = 330 ms−1

(a) Frequency of whistle n0 = 1600 Hz
Velocity of source vs = 12 km/h = 12×518=103 ms-1
Velocity of an observer v0 = 0 ms−1
Frequency of whistle received by wall n = ?
Frequency of sound received by the observer is given by:

n = v + v0v – vs×n0

On substituting the respective values in the above formula, we get:

n=330+0330-103×1600=1616 Hz
(b) Here,
Velocity of observer v0 = 103 ms-1
Velocity of source vs = 0
Frequency of source n0 = 1616 Hz
Frequency of sound heard by observer is
n=v+v0v+vs×n0

On substituting the respective values in the above formula, we get:

=330+103330+0×1616=1632 Hz


Answer: 82

Given:
Velocity of sound in air v = 330 ms−1
Frequency of signal emitted by the source n0 = 1600 Hz

Velocity of source vs = 72 kmh−1 = 72×518 = 20 ms-1

As the sound gets reflected, therefore:

Velocity of source ( vs ) = Velocity of observer ( vL )

Velocity of sound heard by the observer is given by:

n=v+vLv-vs×n0

On substituting the values, we get:

n = 330-20330+20×1600=1417 Hz

The frequency of the reflected signal as heard by the person is 1417 Hz.


Answer: 83

Given:
Velocity of car vcar = 54 kmh−1 = 54×518=15 ms-1
Frequency of the car f = 400 Hz
Velocity of sound in air vair = 335 ms−1
Wavelength in front of the car λ = ?

(a) Net velocity in front of the car v = vcar-vair = 335-15 = 320 m/s

As v=fλ,∴ λ=vf
⇒λ=320400=80 cm

(b) The frequency f1 heard near the cliff is given by:

f1=vairvair+vcar×f0⇒f1=335335+5×400⇒f1=335×400320 Hz⇒f1=418.75 Hz

As we know,
v=fλ.

Wavelength reflected from the cliff is λ=vf1=335418.75=80 cm

(c) Here, v0 = 15 ms-1.

Frequency of the reflected sound wave f2 heard by the person sitting in the car:

f2=v+v0v×f1⇒f2=335+15335×335320×400⇒f2=437 Hz

(d) He will not hear any beat in 10 seconds because the difference of frequencies is greater than 10 (persistence of sound for the human ear is 1/10 of a second).


Answer: 84

Given:
Velocity of sound in air v = 324 ms−1
Frequency of sound sent by source n0 = 400 Hz
Let the speed of the car be x m/s.
The frequency of sound heard at the car n is given by:
n=v+vcarv×n0⇒ n=324+x324×400 …..1
If n1 is the frequency of sound heard by the operator, then its value is given by:

n1=324324-x×n

410=324324-x×n

On substituting the value of n from equation (1), we have:

410=324324-x×324+x324×400⇒ 410=324+x324-x×400⇒ 410 324-x=400324+x⇒ 324 410-400=810x⇒ x=4 m/s

The speed of the car is 4 m/s.


Answer: 85

Given:
Velocity of sound in air v = 330 ms−1
Distance travelled by the sound s = 330 m
Frequency of the sound n = 2 kHz
(a) Velocity v = st

∴ Time t = 330330 = 1 s

(b) The frequency of sound heard by the listener is 2 kHz.
(Since frequency does not depend on distance.)

(c) s = 22 m (= 22 m/s × 1 s) away from P on x-axis.


Answer: 86

Given:
Speed of sound in air v = 330 ms−1
Frequency of sound f0 = 4000 Hz
Velocity of source vs = 22 m/s
The apparent frequency heard by the listener f = ?

At t = 0, let the source be at a distance of y from the origin. Now, the time taken by the sound
to reach the listener is the same as the time taken by the sound to reach the origin.
∴​  y22=660+y2330⇒ 15y2=6602+y2⇒ 224y2=6602⇒y=660224

Velocity of source along the line joining the source S and listener L:
vscosθ = 22.y660+y2=22y15y=2215

Frequency heard by the listener f is

f=vv-vs cosθ×f0⇒f=330330-2215×4000

⇒f= 4017.85 ≈ 4018 Hz


Answer: 87

Given:

Velocity of the source vs = 170 m/s
Frequency of the source f0 = 1200 Hz
(a) 

As shown in the figure,
the time taken by the sound to reach the listener is the same as the time taken by the sound to reach the point of intersection.

y170=2002+y2340⇒ 2y2=2002+y2⇒ 4y2-y2=2002⇒3y2=2002⇒y=2003
Frequency of source will be:
vscosθ = 170.y2002+y2=170×12=85

The frequency of sound f heard by the detector is given by:

f=vv-vs cos v×f0⇒f=340340-170×12×1200⇒f=1600 Hz

(b) The detector will detect a frequency of 1200 Hz at a minimum distance.

200340=x170⇒ x=100 m

∴ Distance

=2002+x2=2002+1002=224 m


Answer: 88

Given:
Frequency of sound emitted by the source f0 = 500 Hz
Velocity of sound in air v = 330 ms-1
Radius of the circle r = 1.6 m

Frequency of sound heard by the observer v, = ?

(a)

Velocity of source at highest point of the circle A is given by:
vs = rg =10×1.6 = 4 m/s

Velocity of sound at C is

vc=5rg=5×1.6×10 =8.9 m/s

The frequency of sound heard by the observer when the source is at point C:

fC=vv – vs×f0

Substituting the values, we get:

fC=330330 – 8.9×500⇒ fC=513.85 Hz≈514 Hz

Frequency of sound observed by the observer when the source is at point A:

fA=vv+vs×n0 =300300+4×500=494 Hz

Therefore, maximum frequency heard by the observer is 514 Hz.

(b) Velocity at B is given by:

vB=3rg=3×1.6×10=6.92 m/s

Frequency at B fB will be:

fB=vv+vs×f0 =330330+6.92×500=490 Hz

Frequency at D fD will be:

fD=vv-vs×f0 =330330-6.92×500=511 Hz


Answer: 89

Let d be the initial distance between the source and the observer.
If v is the speed of sound emitted by the observer, then the time taken by the sound to reach the observer is given by:
T1 = d/v
The source is also moving. Therefore, at t = T, it moves a distance of (s) and is given by:
s=0×T+12aT2

Time taken by the pulse to reach the observer:
d-12aT2v

Time difference ∆t between the two pulses:

T+d-12aT2v-dv =T-aT22v

On replacing u = 1T,

the apparent frequency will be:

1∆t= 2uv22uv-a.


 

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