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The frequency of the source, ν = 500 Hz. 

The speed of the source uₛ = 90 km/h 

=90000/3600 m/s =25 m/s 

The speed of the observer, uₒ =90 km/h =25 m/s. 

The speed of the sound V = 350 m/s. 

Hence the apparent frequency heard in the other train ν' = (V+uₒ)ν/(V-uₛ) 

=(350+25)*500/(350-25) 

=375*500/325 Hz 

=577 Hz

The frequency of sound, ν = 16 kHz. V = 330 m/s. The apparent frequency (minimum for beyond the hearing range) ν' = 20 kHz. 

The speed of observer =? 

ν' = (V+u)ν/V 

→20 = (330+u)*16/330 

→330+u = 20*330/16 =412.5 

→u = 82.5 m/s = 82.5*3600/1000 km/h 

→u = 297 km/h  

Hence if the claim of the driver is assumed to be true, then he must be driving the car more than 297 km/h ≈300 km/h 

In today's technology, general make of cars and road conditions, it is not practical.  

The correct answer is (d) = v.

EXPLANATION: 

Since the source and the observer both move with the same speed there is no relative motion. Thus there is no Doppler's effect and no change in apparent frequency. Hence the option (d).

(a) The distance of the listener from the origin, d = 330 m. The speed of sound in air, V = 330 m/s. The time taken by the sound to reach the listener, 

t = d/V =330/330 s = 1 second.  

(b) Since the direction of the movement of the source is perpendicular to the line joining the source and the person, hence the speed of the source towards the listener will be zero. The same frequency of 2.0 kHz will be heard by the person.   

(c) Since the time to reach the listener = 1 s and the speed of the source = 22 m/s. Hence at this instant, the source will move 22 m. Its position will be  at x = 22 m.

The distance between two consecutive nodes is equal to half of the wavelength of the vibration. This distance given here = 4.0 cm =0.04 m. 

Hence /2 = 0.04 m 

→ = 2*0.04 m =0.08 m 

Since the speed of sound in air, V = 328 m/s, so the frequency of the source, ν = V/ 

→ν = 328/0.08 Hz =32800/8 Hz  

→ν = 4100 Hz = 4.10 kHz.

The resonant frequencies in an open organ pipe is given as ν = nV/2L. Where n = 1, 2, 3, ... 

Given V = 340 m/s and L = 50 cm =0.50 m. 

Hence, 

ν = n(340/2*0.50) =340 n. 

The resonant frequencies between 1000 Hz to 2000 Hz can be found by putting n = 3, 4 and 5 in it.  

For n = 3, ν₃ = 340*3 = 1020 Hz, 

for n = 4, ν₄ = 340*4 = 1360 Hz, 

for n = 5, ν₅ = 340*5 = 1700 Hz.

The separation between a node and the next antinode is equal to one-fourth of the wavelength. Hence from the given data of the problem, 

λ/4 = 25 cm = 0.25 m 

→ λ= 4*0.25 m =1.0 m. 

The speed of sound in air V = 340 m/s. 

Hence the frequency of the vibration ν = V/λ

→ν = 340/1.0 Hz =340 Hz

The loudspeakers have very high intensity in the front direction but not so high in the backward direction. In the given condition let us assume that both loudspeakers are connected to the same source. The person behind the first loudspeaker will hear comparatively very low-intensity sound from it than the other one if the distance between the loudspeakers is small. Even if their sound interferes either constructively or destructively at the ears of the person, he will hear the sound clearly. But if the distance between the loudspeakers is large, the intensity of the other loudspeaker will be comparable to the first one, also due to the path difference both sources will have a phase difference and the clarity of the sound will be seriously affected.

(a)The frequency of the sound, ν = 2000 Hz 

Speed of sound V = 1500 m/s.  

The speed of the source uₛ = 36 km/h 

=36000/3600 m/s =10 m/s 

The speed of the observer uₒ = 54 km/h 

=54000/3600 m/s 

=30/2 m/s =15 m/s 

Hence the apparent frequency received by the second submarine

 ν' = (V+uₒ)ν/(V-uₛ) 

=(1500+15)2000/(1500-10) 

=1515*2000/1490 Hz 

=2034 Hz  

(b) This apparent frequency is now a source when the signal is reflected. The frequency of the sound received by the first submarine 

ν" =(V+uₒ)ν/(V-uₛ) 

But now uₒ = 10 m/s and uₛ = 15 m/s and ν = ν' =2034 Hz 

Hence, ν"=(1500+10)*2034/(1500-15) 

=1510*2034/1485 

=2068 Hz

(c) v₂ > v₂ > v₁  

EXPLANATION: 

When the source is at C the source speed is perpendicular to the line joining the observer and the source. Thus at this instant, the separation between the two is not changing and the frequency heard ν₃ is same as the source. When the source is at A the separation between them is increasing, so due to the Doppler effect ν₁ < ν₃. But when the source is at B, the separation is decreasing and due to the Doppler effect ν₃ < ν₂. So ν₂ > ν₃ > ν₁. Hence the option (c).

(d) Wave velocity.

EXPLANATION: 

When we speak there are different frequencies, wavelength, and amplitudes in the sound produced. The frequency and the wavelength changes with the pitch of the sound and the amplitude vary with the loudness. But the wave velocity remains the same in the air when the temperature is constant. Option (d) is true.

(d) separation between the source and the observer.

EXPLANATION: 

The apparent frequency due to the Doppler effect ν' = {V/(V-U)}*ν₀  or ν' = {(V+U)/V}*ν₀ depending upon the source moves towards the observer or the observer moves towards the source. Thus the change in frequency due to Doppler effect depends on V, U and ν₀ only. It does not depend on the separation between the source and the observer. So the option (d).

Speed of the source u = 220 m/s. The speed of sound in air = 330 m/s. Let the original frequency =ν.  

When the bullet is approaching, the source is approaching and the observer is stationary, the apparent frequency 

ν₁ =Vν/(V-u) =330ν/(330-220) 

→ν₁ = 330ν/110 =3ν 

When the bullet has crossed, the source is receding and the observer is stationary. The apparent frequency now,

 ν₂ = Vν/(V+u) =330ν/(330+220) 

→ν₂ = 330ν/550 =3ν/5 

The change in apparent frequency = ν₁-ν₂ =3ν-3ν/5 = 12ν/5 

Hence the fractional change in apparent frequency with respect to initial frequency = (Change in apparent frequency)÷(Initial apparent frequency) 

=(12ν/5)÷(3ν) 

=4/5 

=0.8

(d) p = ∑ p_on sin(k_nx - ω_nt).

EXPLANATION: 

When we clap, different regions of both the palms do not strike each other with the same pressure. Hence different regions of the palms produce different sounds having different pressure amplitudes, wave numbers and frequencies. So the pressure at a point is given as summation of pressures at that point at that instant. Hence the option (d). 

The first overtone frequency of the closed organ pipe ν =3V/4L₁ 

The fundamental frequency of an open organ pipe  ν =V/2L₂ 

Equating, 3V/4L₁ = V/2L₂ 

→L₂ = 2L₁/3            {Given L₁ = 30 cm} 

→L₂ = 2*30/3 cm =20 cm.

The resonant frequencies in an open organ pipe is given as, 

νₙ = nV/2L, where V = 340 m/s, L = 20 cm =0.20 m 

→νₙ = n*(340/2*0.20) =n*3400/4 =n*850 Hz. For the fundamental frequency, n = 1, ν₀ = 1*850 Hz =850 Hz  

For the first overtone frequency, n = 2, 

ν₁ = 2*850 Hz = 1700 Hz  

For the second overtone frequency, n = 3, 

ν₂ = 3*850 Hz =2550 Hz

The resonant frequencies in an open organ pipe is given as  

νₙ = nV/2L 

Let the given frequencies are for n and n+1 harmonics, thus 

nV/2L = 1944 and  

(n+1)V/2L = 2592 

→nV/2L + V/2L =2592 

→1944+V/2L = 2592 

→V/2L = 2592 - 1944 = 648

 →L = V/(2*648) = 324/1296 m = 0.25 m

 →L = 25 cm

(a) longitudinal stationary waves

EXPLANATION:

 An organ pipe open at both ends contains sound waves which are longitudinal waves, so the option (c) and (d) are not true. When the waves enter through one end it gets reflected from the other end with a phase change of π. The reflected wave is again reflected by the first end. With repeated reflections, if the length of the pipe is a multiple of the half wavelength, stationary waves are formed. Thus option (a).

Frequency of the source, ν = 2400 Hz. Speed of the source, u = 18.0 km/h =18000/3600 m/s = 5 m/s 

Speed of sound in air, V = 340 m/s.  

Here the source is approaching the observer and the observer is stationary. Hence the apparent frequency for the observer, ν' = Vν/(V-u) 

=340*2400/(340-5) Hz 

=2436 Hz

(a) The frequency of the sound ν = 500 Hz. The speed of sound in air V = 340 m/s. Speed of the source uₛ = 108 km/h =108000/3600 m/s =30 m/s. 

The speed of the observer uₒ = 108 km/k =30 m/s. 

In such case apparent frequency ν' = (V+uₒ)ν/(V-uₛ) 

But here the source is leaving hence 

ν' = (V+uₒ)ν/(V+uₛ) = ν = 500 Hz {since uₒ = uₛ} 

It can also be concluded by the fact that the relative motion between the source and the observer is zero.  

(b) For the person standing near the track uₒ = 0 and the source is leaving. 

ν' = Vν/(V+uₛ) =340*500/(340+30) Hz  =340*500/370 Hz 

=459 Hz  

(c) The speed of the medium uₘ =36 km/h 

=36000/3600 m/s = 10 m/s towards east. 

Hence the effective speed of source = uₛ+uₘ and the effective speed of the passenger = uₒ+uₘ Both are in the same direction hence no relative motion. Thus the frequency observed by the passenger = ν = 500 Hz.  

The effective speed of the sound in the air for the person standing due to the wind towards the east = V-uₘ =340-10 =330 m/s. Now the apparent frequency for the observer can be calculated as usual 

ν' = Vν/(V+uₛ) 

=330*500/(330+30) Hz 

=330*500/360 Hz 

=458 Hz

(a) The apparent frequency heard by the standing person will be more than 440 Hz due to Doppler's effect. Since he hears 4 beats per second, the apparent frequency of the violin in the train, ν' = 440+4 Hz =444 Hz. 

Here ν =440 Hz 

V = 340 m/s 

u =? ν' 

= Vν/(V-u) 

→444 =340*440/(340-u) 

→340-u = 340*440/444 =336.94 

→u = 340 - 336.94 m/s = 3.06 m/s 

 =3.06*3600/1000 km/h ≈11 km/h 

 (b) For the person in the train, the source on the ground is stationary and observer approaches. The apparent frequency heard by the person in the train ν'' =ν(V+u)/V 

= 440*(340+3.06)/340 Hz  

=440*1.009 Hz 

=443.96 Hz 

The beat frequency heard when the train approaches =443.96 - 440 = 3.96 beats/s. i.e. a little less than 4 beats/s

Given that V = 330 m/s, the air columns in the U-tube are the closed organ pipes. The fundamental frequency  

V/4L = 440 Hz 

→L = V/(4*440) m = 330/(4*440) m 

→L = 3/16 m = 0.188 m = 18.8 cm  

The first overtone frequency is  

3V/4L = 440 Hz

 →L = 3V/(4*440) m =3*330/(4*440) m 

→L = 9/16 m = 0.563 m = 56.3 cm

 The fundamental frequency of the closed organ pipe at 20°C = ν₀ and at 22°C = ν₁. Since the frequency is proportional to the speed of sound V and the speed of sound is proportional to the square root of the temperature, hence ν₀/ν₁ =√(T₀/T₁) 

Given ν₀ = 293 Hz, T₀ = 273+20 =293 K, T₁ = 273+22 =295 K. 

→293/ν₁ = √(293/295) 

→ν₁ = 293*√(295/293) =√(295*293) 

=294 Hz