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SOUND WAVES
EXERCISES- Q-21 to Q-30
21. Sound with intensity larger than 120 dB appears painful to a person. A small speaker delivers 2.0 W of audio output. How close can the person get to the speaker without hurting his ears?
ANSWER: The intensity of the sound at a distance r from the speaker I = 2.0/4πr² =1/2πr²
The sound level at this distance = 10 log [I/I₀]
The person can get close up to the distance where the sound level is 120 dB. Let this distance be r, then,
120 = 10 log [(1/2πr²)/10⁻¹²]
{Since the reference intensity I₀ = 10⁻¹² W/m²}
→log [10¹²/2πr²] = 12
→log 10¹² -log [2πr²] = 12
→12 - log[2π] - log r² = 12
→log r² = -log [2π] = log [2π]⁻¹
→r² = [2π]⁻¹ =1/2π =0.16
→r = 0.40 m = 40 cm
22. If the sound level in a room is increased from 50 dB to 60 dB, by what factor is the pressure amplitude increased?
ANSWER: Let the intensity of sound at 50 dB = I and at 60 dB = I'. Hence
10 log[I/I₀] = 50 and
10 log[I'/I₀] = 60
Subtracting the first from the second,
10 log[I'/I₀] - 10 log[I/I₀] = 60-50 = 10
→log[I'/I] = 1
→I'/I = 10
The relation between the intensity and the pressure amplitude is given as
I = p₀²/2ρv {where ρ is the density of the medium and v is the speed of the sound wave in the medium. Both are constant here}
Hence I'/I = p₀'²/p₀²
→(p₀'/p₀)² =I'/I =10
→p₀'/p₀ = √10
23. The noise level in a classroom in absence of the teacher is 50 dB when 50 students are present. Assuming that on the average each student outputs same sound energy per second, what will be the noise level if the number of students is increased to 100?
ANSWER: The intensity of the sound will be doubled in the given condition. Thus,
I'/I = 2
If the sound level in the second condition is ß, then
ß - 50 = 10 log[I'/I₀] - 10 log[I/I₀]
→ß - 50 = 10 log[I'/I] =10 log[2] ≈10*0.3 =3
→ß = 50 + 3 = 53 dB
24. In a Quincke's experiment, the sound detected is changed from a maximum to minimum when the sliding tube is moved through a distance of 2.50 cm. Find the frequency of sound if the speed of sound in air is 340 m/s.
ANSWER: When the tube is moved by 2.50 cm the path length changes to 2*2.50 cm = 5.0 cm.
Since sound detected changes from maximum to minimum the waves interfere destructively. Here for the consecutive maximum and minimum the phase difference is the least odd multiple of π i.e. π. So,
ẟ = 2π*5.0/𝜆 =π
→𝜆 = 10 cm =0.10 m
Hence the frequency of the sound
ν = V/𝜆 = 340/0.10 Hz = 3400 Hz =3.4 kHz
25. In a Quincke's experiment, the sound intensity has a minimum value I at a particular position. As the sliding tube is pulled out by a distance of 16.5 mm, the intensity increases to a maximum of 9I. Take the speed of sound in air to be 330 m/s. (a) Find the frequency of the sound source. (b) Find the ratio of the amplitudes of the two waves arriving at the detector assuming that it does not change much between the positions of minimum intensity and maximum intensity.
ANSWER: (a) In this case the path difference
= 16.5*2 mm = 33 mm =0.033 m
Since the intensity changes from minimum to maximum, the phase difference (ẟ) is π.
ẟ = 2π*0.033/𝜆
→π = 2π*0.033/𝜆
→𝜆 = 0.066
Hence the frequency of the sound,
ν = V/𝜆 = 330/0.066 =5000 Hz =5.0 kHz
(b) Let p₀' = the amplitude of the first wave and p₀ of the second. Thus p₀'/ p₀ = K =?
The amplitude at the minimum intensity =p₀' - p₀
The amplitude at the maximum intensity =p₀' + p₀
Since the intensity is proportional to the square of the pressure amplitudes, hence
9I/I = (p₀' + p₀)²/(p₀' - p₀)²
→9 = (p₀'/ p₀ + 1)²/(p₀'/p₀ - 1)²
→(K+1)² = 9(K-1)²
→K+1 = 3(K-1) =3K-3
→3K-K = 3+1
→2K = 4
→K = 2
26. Two audio speakers are kept some distance apart and are driven by the same amplifier system. A person is sitting at a place 6.0 m from one of the speakers and 6.4 m from the other. If the sound signal is continuously varied from 500 Hz to 5000 Hz, what are the frequencies for which there is a destructive interference at the place of the listener? The speed of sound in air = 320 m/s.
ANSWER: The destructive interference will be for the phase difference between the two waves having odd multiple of π.
The path difference =6.4 - 6.0 =0.40 m
The phase difference
ẟ = 2π*0.40/𝜆 =2π*0.40*ν/V
→ẟ = 2π*0.40*ν/320 =π*{0.0025 ν}
So for the ẟ to be an odd multiple of π, the factor {0.0025ν} must be an odd number.
i.e. 0.0025ν = 1, 3, 5, 7, 9, 11, 13, .....
→ν/400 = 1, 3, 5, 7, 9, 11, 13, .....
→ν = 400 Hz, 1200 Hz, 2000 Hz, 2800 Hz, 3600 Hz, 4400 Hz, 5200 Hz, ....
But the given frequency range is between 500 Hz to 5000 Hz. Hence the destructive interference at the place for the given frequency range is for the frequencies 1200 Hz, 2000 Hz, 2800 Hz, 3600 Hz, 4400 Hz.
27. A source of sound S and a detector D are placed at some distance from one another. A big cardboard is placed near the detector and perpendicular to the line SD as shown in figure (16-E1). It is gradually moved away and it is found that the intensity changes from a maximum to minimum as the board is moved through a distance of 20 cm. Find the frequency of the sound emitted. The velocity of the sound in air is 336 m/s.
Figure for Q - 27
ANSWER: The reflected sound waves from the cardboard have a path difference = 2*20 cm =40 cm =0.40 m.
Hence the phase difference ẟ =2π*0.40/𝜆
→ẟ = π*(0.80 ν/V) =π*(0.80 ν/336)
Since the intensity changes from maximum to minimum, the phase difference is π.
Thus the phase difference
π*(0.80 ν/336) = π
→0.80 ν/336 = 1
→ν = 336/0.80 Hz = 420 Hz
28. A source S and a detector D are placed at a distance d apart. A big cardboard is placed at a distance H from the source and the detector as shown in the figure (16-E2). The source emits a wave of wavelength = d/2 which is received by the detector after the reflection from the cardboard. It is found to be in phase with the direct wave received from the source. By what minimum distance should the cardboard be shifted away so that the reflected wave becomes out of phase with the direct wave?
Diagram for Q - 28
ANSWER: Let the required minimum distance for the shift of the cardboard = x. (see the diagram below).
Diagram for Q - 28
The path difference between the present and the previous reflected waves
Δx = SO'D - SOD =2(SO'-SO)
=2{√(SM² + MO'²) - √(SM²+MO²)}
=2{√(d²/4 + (√2d + x)²) - √(d²/4 + 2d²)}
=2{√(d²/4 + 2d²+x²+2√2xd) - √(d²/4 +2d²)}
neglecting x²
Δx = 2{√(d²/4 +2d²+2√2xd) - √(d²/4 +2d²)}
Since the reflected wave just becomes out of phase,
Δx =𝜆/2 = d/4, thus
2{√(d²/4 +2d²+2√2xd) - √(d²/4 +2d²)} =d/4
→√(9d²/4 +2√2xd) - √(9d²/4) = d/8
→√(9d²/4 +2√2xd) = √(9d²/4) + d/8
squaring,
→9d²/4 +2√2xd = 9d²/4 + d²/64 +2(d/8)(3d/2)
→2√2xd = d²/64 +3d²/8 =25d²/64
→x = 25d/(64*2√2) =0.13 d
29. Two stereo speakers are separated by a distance of 2.40 m. A person stands at a distance of 3.20 m directly in front of one of the speakers as shown in figure (16-E3). Find the frequencies in the audible range (20-20000 Hz) for which the listener will hear a minimum sound intensity. The speed of sound in air = 320 m/s.
Figure for Q - 29
ANSWER: The phase difference ẟ =2π*Δx/𝜆
ẟ =2π*Δx*ν/V =π*(2Δx*ν/V)
Where Δx = path difference, V =speed of sound =320 m/s, and ν = frequency
Here Δx =√(3.2²+2.4²) - 3.2 m =4-3.2 m =0.80 m
Hence ẟ =π*(2*0.80*ν/320) =π*(ν/200)
For the listener to hear minimum intensity, the phase difference ẟ must be an odd multiple of π. i.e.
ν/200 =(2n+1), where n = 0,1,2, ....
→ν = 200(2n+1) Hz, where n = 0,1,2, ....
For the ν to be in the audible range (200 Hz to 20000 Hz), n =0, 1, 2, 3, .... 49. Because for n = 50, ν =20200 Hz. Hence the required frequencies are
ν = 200(2n+1) Hz where n = 0,1,2, ..... 49
30. Two sources of sound, S₁ and S₂, emitting waves of equal wavelength 20.0 cm, are placed with a separation of 20.0 cm between them. A detector can be moved on a line parallel to S₁S₂ and at a distance of 20.0 cm from it. Initially, the detector is equidistant from the two sources. Assuming that the waves emitted by the sources are in phase, find the minimum distance through which the detector should be shifted to detect a minimum of sound.
ANSWER: Let the minimum distance shifted be x as in the diagram below.
Diagram for Q-30
The path difference Δx = ES₁ -ES₂
→Δx = √(S₁G²+EG²) -√(S₂G²+EG²)
→Δx = √{(10+x)²+20²} - √{(10-x)²+20²}
The phase difference ẟ = 2π*Δx/𝜆
→ẟ = 2π[√{(10+x)²+20²} - √{(10-x)²+20²}]/20
→ẟ = π*[√{(10+x)²+20²} - √{(10-x)²+20²}]/10
For the given condition ẟ =π, hence
[√{(10+x)²+20²} - √{(10-x)²+20²}]/10 =1
→√{(10+x)²+20²} - √{(10-x)²+20²} = 10
Squaring both sides,
{(10+x)²+20²} + {(10-x)²+20²} -2√{(10+x)²+20²}*√{(10-x)²+20²} = 100
→100+x²+20x+100+x²-20x +800-2√{(100-x²)²+20²(200+2x²)+400²} =100
→200+2x²+800-2√{100²+(x²)²-200x²+200*20²+800x²+400²} =100
→x²+450 = √{x⁴+600x²+400²+100²+200*20²}
Again squaring,
x⁴+900x²+450² =x⁴+600x²+400²+100²+200*20²
→300x² = 400²+100²+200*20²-450²
→x² =(160000+10000+80000-202500)/300
→x² =47500/300 =158.33
→x =12.6 cm
21. Sound with intensity larger than 120 dB appears painful to a person. A small speaker delivers 2.0 W of audio output. How close can the person get to the speaker without hurting his ears?
ANSWER: The intensity of the sound at a distance r from the speaker I = 2.0/4πr² =1/2πr²
The sound level at this distance = 10 log [I/I₀]
The person can get close up to the distance where the sound level is 120 dB. Let this distance be r, then,
120 = 10 log [(1/2πr²)/10⁻¹²]
{Since the reference intensity I₀ = 10⁻¹² W/m²}
→log [10¹²/2πr²] = 12
→log 10¹² -log [2πr²] = 12
→12 - log[2π] - log r² = 12
→log r² = -log [2π] = log [2π]⁻¹
→r² = [2π]⁻¹ =1/2π =0.16
→r = 0.40 m = 40 cm
22. If the sound level in a room is increased from 50 dB to 60 dB, by what factor is the pressure amplitude increased?
ANSWER: Let the intensity of sound at 50 dB = I and at 60 dB = I'. Hence
10 log[I/I₀] = 50 and
10 log[I'/I₀] = 60
Subtracting the first from the second,
10 log[I'/I₀] - 10 log[I/I₀] = 60-50 = 10
→log[I'/I] = 1
→I'/I = 10
The relation between the intensity and the pressure amplitude is given as
I = p₀²/2ρv {where ρ is the density of the medium and v is the speed of the sound wave in the medium. Both are constant here}
Hence I'/I = p₀'²/p₀²
→(p₀'/p₀)² =I'/I =10
→p₀'/p₀ = √10
23. The noise level in a classroom in absence of the teacher is 50 dB when 50 students are present. Assuming that on the average each student outputs same sound energy per second, what will be the noise level if the number of students is increased to 100?
ANSWER: The intensity of the sound will be doubled in the given condition. Thus,
I'/I = 2
If the sound level in the second condition is ß, then
ß - 50 = 10 log[I'/I₀] - 10 log[I/I₀]
→ß - 50 = 10 log[I'/I] =10 log[2] ≈10*0.3 =3
→ß = 50 + 3 = 53 dB
24. In a Quincke's experiment, the sound detected is changed from a maximum to minimum when the sliding tube is moved through a distance of 2.50 cm. Find the frequency of sound if the speed of sound in air is 340 m/s.
ANSWER: When the tube is moved by 2.50 cm the path length changes to 2*2.50 cm = 5.0 cm.
Since sound detected changes from maximum to minimum the waves interfere destructively. Here for the consecutive maximum and minimum the phase difference is the least odd multiple of π i.e. π. So,
ẟ = 2π*5.0/𝜆 =π
→𝜆 = 10 cm =0.10 m
Hence the frequency of the sound
ν = V/𝜆 = 340/0.10 Hz = 3400 Hz =3.4 kHz
25. In a Quincke's experiment, the sound intensity has a minimum value I at a particular position. As the sliding tube is pulled out by a distance of 16.5 mm, the intensity increases to a maximum of 9I. Take the speed of sound in air to be 330 m/s. (a) Find the frequency of the sound source. (b) Find the ratio of the amplitudes of the two waves arriving at the detector assuming that it does not change much between the positions of minimum intensity and maximum intensity.
ANSWER: (a) In this case the path difference
= 16.5*2 mm = 33 mm =0.033 m
Since the intensity changes from minimum to maximum, the phase difference (ẟ) is π.
ẟ = 2π*0.033/𝜆
→π = 2π*0.033/𝜆
→𝜆 = 0.066
Hence the frequency of the sound,
ν = V/𝜆 = 330/0.066 =5000 Hz =5.0 kHz
(b) Let p₀' = the amplitude of the first wave and p₀ of the second. Thus p₀'/ p₀ = K =?
The amplitude at the minimum intensity =p₀' - p₀
The amplitude at the maximum intensity =p₀' + p₀
Since the intensity is proportional to the square of the pressure amplitudes, hence
9I/I = (p₀' + p₀)²/(p₀' - p₀)²
→9 = (p₀'/ p₀ + 1)²/(p₀'/p₀ - 1)²
→(K+1)² = 9(K-1)²
→K+1 = 3(K-1) =3K-3
→3K-K = 3+1
→2K = 4
→K = 2
26. Two audio speakers are kept some distance apart and are driven by the same amplifier system. A person is sitting at a place 6.0 m from one of the speakers and 6.4 m from the other. If the sound signal is continuously varied from 500 Hz to 5000 Hz, what are the frequencies for which there is a destructive interference at the place of the listener? The speed of sound in air = 320 m/s.
ANSWER: The destructive interference will be for the phase difference between the two waves having odd multiple of π.
The path difference =6.4 - 6.0 =0.40 m
The phase difference
ẟ = 2π*0.40/𝜆 =2π*0.40*ν/V
→ẟ = 2π*0.40*ν/320 =π*{0.0025 ν}
So for the ẟ to be an odd multiple of π, the factor {0.0025ν} must be an odd number.
i.e. 0.0025ν = 1, 3, 5, 7, 9, 11, 13, .....
→ν/400 = 1, 3, 5, 7, 9, 11, 13, .....
→ν = 400 Hz, 1200 Hz, 2000 Hz, 2800 Hz, 3600 Hz, 4400 Hz, 5200 Hz, ....
But the given frequency range is between 500 Hz to 5000 Hz. Hence the destructive interference at the place for the given frequency range is for the frequencies 1200 Hz, 2000 Hz, 2800 Hz, 3600 Hz, 4400 Hz.
27. A source of sound S and a detector D are placed at some distance from one another. A big cardboard is placed near the detector and perpendicular to the line SD as shown in figure (16-E1). It is gradually moved away and it is found that the intensity changes from a maximum to minimum as the board is moved through a distance of 20 cm. Find the frequency of the sound emitted. The velocity of the sound in air is 336 m/s.
ANSWER: The reflected sound waves from the cardboard have a path difference = 2*20 cm =40 cm =0.40 m.
Hence the phase difference ẟ =2π*0.40/𝜆
→ẟ = π*(0.80 ν/V) =π*(0.80 ν/336)
Since the intensity changes from maximum to minimum, the phase difference is π.
Thus the phase difference
π*(0.80 ν/336) = π
→0.80 ν/336 = 1
→ν = 336/0.80 Hz = 420 Hz
28. A source S and a detector D are placed at a distance d apart. A big cardboard is placed at a distance H from the source and the detector as shown in the figure (16-E2). The source emits a wave of wavelength = d/2 which is received by the detector after the reflection from the cardboard. It is found to be in phase with the direct wave received from the source. By what minimum distance should the cardboard be shifted away so that the reflected wave becomes out of phase with the direct wave?
ANSWER: Let the required minimum distance for the shift of the cardboard = x. (see the diagram below).
The path difference between the present and the previous reflected waves
Δx = SO'D - SOD =2(SO'-SO)
=2{√(SM² + MO'²) - √(SM²+MO²)}
=2{√(d²/4 + (√2d + x)²) - √(d²/4 + 2d²)}
=2{√(d²/4 + 2d²+x²+2√2xd) - √(d²/4 +2d²)}
neglecting x²
Δx = 2{√(d²/4 +2d²+2√2xd) - √(d²/4 +2d²)}
Since the reflected wave just becomes out of phase,
Δx =𝜆/2 = d/4, thus
2{√(d²/4 +2d²+2√2xd) - √(d²/4 +2d²)} =d/4
→√(9d²/4 +2√2xd) - √(9d²/4) = d/8
→√(9d²/4 +2√2xd) = √(9d²/4) + d/8
squaring,
→9d²/4 +2√2xd = 9d²/4 + d²/64 +2(d/8)(3d/2)
→2√2xd = d²/64 +3d²/8 =25d²/64
→x = 25d/(64*2√2) =0.13 d
29. Two stereo speakers are separated by a distance of 2.40 m. A person stands at a distance of 3.20 m directly in front of one of the speakers as shown in figure (16-E3). Find the frequencies in the audible range (20-20000 Hz) for which the listener will hear a minimum sound intensity. The speed of sound in air = 320 m/s.
ANSWER: The phase difference ẟ =2π*Δx/𝜆
ẟ =2π*Δx*ν/V =π*(2Δx*ν/V)
Where Δx = path difference, V =speed of sound =320 m/s, and ν = frequency
Here Δx =√(3.2²+2.4²) - 3.2 m =4-3.2 m =0.80 m
Hence ẟ =π*(2*0.80*ν/320) =π*(ν/200)
For the listener to hear minimum intensity, the phase difference ẟ must be an odd multiple of π. i.e.
ν/200 =(2n+1), where n = 0,1,2, ....
→ν = 200(2n+1) Hz, where n = 0,1,2, ....
For the ν to be in the audible range (200 Hz to 20000 Hz), n =0, 1, 2, 3, .... 49. Because for n = 50, ν =20200 Hz. Hence the required frequencies are
ν = 200(2n+1) Hz where n = 0,1,2, ..... 49
30. Two sources of sound, S₁ and S₂, emitting waves of equal wavelength 20.0 cm, are placed with a separation of 20.0 cm between them. A detector can be moved on a line parallel to S₁S₂ and at a distance of 20.0 cm from it. Initially, the detector is equidistant from the two sources. Assuming that the waves emitted by the sources are in phase, find the minimum distance through which the detector should be shifted to detect a minimum of sound.
ANSWER: Let the minimum distance shifted be x as in the diagram below.
The path difference Δx = ES₁ -ES₂
→Δx = √(S₁G²+EG²) -√(S₂G²+EG²)
→Δx = √{(10+x)²+20²} - √{(10-x)²+20²}
The phase difference ẟ = 2π*Δx/𝜆
→ẟ = 2π[√{(10+x)²+20²} - √{(10-x)²+20²}]/20
→ẟ = π*[√{(10+x)²+20²} - √{(10-x)²+20²}]/10
For the given condition ẟ =π, hence
[√{(10+x)²+20²} - √{(10-x)²+20²}]/10 =1
→√{(10+x)²+20²} - √{(10-x)²+20²} = 10
Squaring both sides,
{(10+x)²+20²} + {(10-x)²+20²} -2√{(10+x)²+20²}*√{(10-x)²+20²} = 100
→100+x²+20x+100+x²-20x +800-2√{(100-x²)²+20²(200+2x²)+400²} =100
→200+2x²+800-2√{100²+(x²)²-200x²+200*20²+800x²+400²} =100
→x²+450 = √{x⁴+600x²+400²+100²+200*20²}
Again squaring,
x⁴+900x²+450² =x⁴+600x²+400²+100²+200*20²
→300x² = 400²+100²+200*20²-450²
→x² =(160000+10000+80000-202500)/300
→x² =47500/300 =158.33
→x =12.6 cm
ANSWER: The intensity of the sound at a distance r from the speaker I = 2.0/4πr² =1/2πr²
The sound level at this distance = 10 log [I/I₀]
The person can get close up to the distance where the sound level is 120 dB. Let this distance be r, then,
120 = 10 log [(1/2πr²)/10⁻¹²]
{Since the reference intensity I₀ = 10⁻¹² W/m²}
→log [10¹²/2πr²] = 12
→log 10¹² -log [2πr²] = 12
→12 - log[2π] - log r² = 12
→log r² = -log [2π] = log [2π]⁻¹
→r² = [2π]⁻¹ =1/2π =0.16
→r = 0.40 m = 40 cm
22. If the sound level in a room is increased from 50 dB to 60 dB, by what factor is the pressure amplitude increased?
ANSWER: Let the intensity of sound at 50 dB = I and at 60 dB = I'. Hence
10 log[I/I₀] = 50 and
10 log[I'/I₀] = 60
Subtracting the first from the second,
10 log[I'/I₀] - 10 log[I/I₀] = 60-50 = 10
→log[I'/I] = 1
→I'/I = 10
The relation between the intensity and the pressure amplitude is given as
I = p₀²/2ρv {where ρ is the density of the medium and v is the speed of the sound wave in the medium. Both are constant here}
Hence I'/I = p₀'²/p₀²
→(p₀'/p₀)² =I'/I =10
→p₀'/p₀ = √10
23. The noise level in a classroom in absence of the teacher is 50 dB when 50 students are present. Assuming that on the average each student outputs same sound energy per second, what will be the noise level if the number of students is increased to 100?
ANSWER: The intensity of the sound will be doubled in the given condition. Thus,
I'/I = 2
If the sound level in the second condition is ß, then
ß - 50 = 10 log[I'/I₀] - 10 log[I/I₀]
→ß - 50 = 10 log[I'/I] =10 log[2] ≈10*0.3 =3
→ß = 50 + 3 = 53 dB
24. In a Quincke's experiment, the sound detected is changed from a maximum to minimum when the sliding tube is moved through a distance of 2.50 cm. Find the frequency of sound if the speed of sound in air is 340 m/s.
ANSWER: When the tube is moved by 2.50 cm the path length changes to 2*2.50 cm = 5.0 cm.
Since sound detected changes from maximum to minimum the waves interfere destructively. Here for the consecutive maximum and minimum the phase difference is the least odd multiple of π i.e. π. So,
ẟ = 2π*5.0/𝜆 =π
→𝜆 = 10 cm =0.10 m
Hence the frequency of the sound
ν = V/𝜆 = 340/0.10 Hz = 3400 Hz =3.4 kHz
25. In a Quincke's experiment, the sound intensity has a minimum value I at a particular position. As the sliding tube is pulled out by a distance of 16.5 mm, the intensity increases to a maximum of 9I. Take the speed of sound in air to be 330 m/s. (a) Find the frequency of the sound source. (b) Find the ratio of the amplitudes of the two waves arriving at the detector assuming that it does not change much between the positions of minimum intensity and maximum intensity.
ANSWER: (a) In this case the path difference
= 16.5*2 mm = 33 mm =0.033 m
Since the intensity changes from minimum to maximum, the phase difference (ẟ) is π.
ẟ = 2π*0.033/𝜆
→π = 2π*0.033/𝜆
→𝜆 = 0.066
Hence the frequency of the sound,
ν = V/𝜆 = 330/0.066 =5000 Hz =5.0 kHz
(b) Let p₀' = the amplitude of the first wave and p₀ of the second. Thus p₀'/ p₀ = K =?
The amplitude at the minimum intensity =p₀' - p₀
The amplitude at the maximum intensity =p₀' + p₀
Since the intensity is proportional to the square of the pressure amplitudes, hence
9I/I = (p₀' + p₀)²/(p₀' - p₀)²
→9 = (p₀'/ p₀ + 1)²/(p₀'/p₀ - 1)²
→(K+1)² = 9(K-1)²
→K+1 = 3(K-1) =3K-3
→3K-K = 3+1
→2K = 4
→K = 2
26. Two audio speakers are kept some distance apart and are driven by the same amplifier system. A person is sitting at a place 6.0 m from one of the speakers and 6.4 m from the other. If the sound signal is continuously varied from 500 Hz to 5000 Hz, what are the frequencies for which there is a destructive interference at the place of the listener? The speed of sound in air = 320 m/s.
ANSWER: The destructive interference will be for the phase difference between the two waves having odd multiple of π.
The path difference =6.4 - 6.0 =0.40 m
The phase difference
ẟ = 2π*0.40/𝜆 =2π*0.40*ν/V
→ẟ = 2π*0.40*ν/320 =π*{0.0025 ν}
So for the ẟ to be an odd multiple of π, the factor {0.0025ν} must be an odd number.
i.e. 0.0025ν = 1, 3, 5, 7, 9, 11, 13, .....
→ν/400 = 1, 3, 5, 7, 9, 11, 13, .....
→ν = 400 Hz, 1200 Hz, 2000 Hz, 2800 Hz, 3600 Hz, 4400 Hz, 5200 Hz, ....
But the given frequency range is between 500 Hz to 5000 Hz. Hence the destructive interference at the place for the given frequency range is for the frequencies 1200 Hz, 2000 Hz, 2800 Hz, 3600 Hz, 4400 Hz.
27. A source of sound S and a detector D are placed at some distance from one another. A big cardboard is placed near the detector and perpendicular to the line SD as shown in figure (16-E1). It is gradually moved away and it is found that the intensity changes from a maximum to minimum as the board is moved through a distance of 20 cm. Find the frequency of the sound emitted. The velocity of the sound in air is 336 m/s.
 |
| Figure for Q - 27 |
ANSWER: The reflected sound waves from the cardboard have a path difference = 2*20 cm =40 cm =0.40 m.
Hence the phase difference ẟ =2π*0.40/𝜆
→ẟ = π*(0.80 ν/V) =π*(0.80 ν/336)
Since the intensity changes from maximum to minimum, the phase difference is π.
Thus the phase difference
π*(0.80 ν/336) = π
→0.80 ν/336 = 1
→ν = 336/0.80 Hz = 420 Hz
28. A source S and a detector D are placed at a distance d apart. A big cardboard is placed at a distance H from the source and the detector as shown in the figure (16-E2). The source emits a wave of wavelength = d/2 which is received by the detector after the reflection from the cardboard. It is found to be in phase with the direct wave received from the source. By what minimum distance should the cardboard be shifted away so that the reflected wave becomes out of phase with the direct wave?
 |
| Diagram for Q - 28 |
ANSWER: Let the required minimum distance for the shift of the cardboard = x. (see the diagram below).
 |
| Diagram for Q - 28 |
The path difference between the present and the previous reflected waves
Δx = SO'D - SOD =2(SO'-SO)
=2{√(SM² + MO'²) - √(SM²+MO²)}
=2{√(d²/4 + (√2d + x)²) - √(d²/4 + 2d²)}
=2{√(d²/4 + 2d²+x²+2√2xd) - √(d²/4 +2d²)}
neglecting x²
Δx = 2{√(d²/4 +2d²+2√2xd) - √(d²/4 +2d²)}
Since the reflected wave just becomes out of phase,
Δx =𝜆/2 = d/4, thus
2{√(d²/4 +2d²+2√2xd) - √(d²/4 +2d²)} =d/4
→√(9d²/4 +2√2xd) - √(9d²/4) = d/8
→√(9d²/4 +2√2xd) = √(9d²/4) + d/8
squaring,
→9d²/4 +2√2xd = 9d²/4 + d²/64 +2(d/8)(3d/2)
→2√2xd = d²/64 +3d²/8 =25d²/64
→x = 25d/(64*2√2) =0.13 d
29. Two stereo speakers are separated by a distance of 2.40 m. A person stands at a distance of 3.20 m directly in front of one of the speakers as shown in figure (16-E3). Find the frequencies in the audible range (20-20000 Hz) for which the listener will hear a minimum sound intensity. The speed of sound in air = 320 m/s.
 |
| Figure for Q - 29 |
ANSWER: The phase difference ẟ =2π*Δx/𝜆
ẟ =2π*Δx*ν/V =π*(2Δx*ν/V)
Where Δx = path difference, V =speed of sound =320 m/s, and ν = frequency
Here Δx =√(3.2²+2.4²) - 3.2 m =4-3.2 m =0.80 m
Hence ẟ =π*(2*0.80*ν/320) =π*(ν/200)
For the listener to hear minimum intensity, the phase difference ẟ must be an odd multiple of π. i.e.
ν/200 =(2n+1), where n = 0,1,2, ....
→ν = 200(2n+1) Hz, where n = 0,1,2, ....
For the ν to be in the audible range (200 Hz to 20000 Hz), n =0, 1, 2, 3, .... 49. Because for n = 50, ν =20200 Hz. Hence the required frequencies are
ν = 200(2n+1) Hz where n = 0,1,2, ..... 49
30. Two sources of sound, S₁ and S₂, emitting waves of equal wavelength 20.0 cm, are placed with a separation of 20.0 cm between them. A detector can be moved on a line parallel to S₁S₂ and at a distance of 20.0 cm from it. Initially, the detector is equidistant from the two sources. Assuming that the waves emitted by the sources are in phase, find the minimum distance through which the detector should be shifted to detect a minimum of sound.
ANSWER: Let the minimum distance shifted be x as in the diagram below.
|
| Diagram for Q-30 |
The path difference Δx = ES₁ -ES₂
→Δx = √(S₁G²+EG²) -√(S₂G²+EG²)
→Δx = √{(10+x)²+20²} - √{(10-x)²+20²}
The phase difference ẟ = 2π*Δx/𝜆
→ẟ = 2π[√{(10+x)²+20²} - √{(10-x)²+20²}]/20
→ẟ = π*[√{(10+x)²+20²} - √{(10-x)²+20²}]/10
For the given condition ẟ =π, hence
[√{(10+x)²+20²} - √{(10-x)²+20²}]/10 =1
→√{(10+x)²+20²} - √{(10-x)²+20²} = 10
Squaring both sides,
{(10+x)²+20²} + {(10-x)²+20²} -2√{(10+x)²+20²}*√{(10-x)²+20²} = 100
→100+x²+20x+100+x²-20x +800-2√{(100-x²)²+20²(200+2x²)+400²} =100
→200+2x²+800-2√{100²+(x²)²-200x²+200*20²+800x²+400²} =100
→x²+450 = √{x⁴+600x²+400²+100²+200*20²}
Again squaring,
x⁴+900x²+450² =x⁴+600x²+400²+100²+200*20²
→300x² = 400²+100²+200*20²-450²
→x² =(160000+10000+80000-202500)/300
→x² =47500/300 =158.33
→x =12.6 cm
===<<<O>>>===
Links to the Chapters
===<<<O>>>===
Links to the Chapters
CHAPTER- 16 - Sound Waves
CHAPTER- 15 - Wave Motion and Waves on a String
OBJECTIVE-II
EXERCISES - Q-1 TO Q-10
EXERCISES - Q-11 TO Q-20
EXERCISES - Q-21 TO Q-30
EXERCISES - Q-31 TO Q-40
EXERCISES - Q-41 TO Q-50
CHAPTER- 14 - Fluid Mechanics
CHAPTER- 13 - Fluid Mechanics
CHAPTER- 12 - Simple Harmonic Motion
EXERCISES- Q1 TO Q10
EXERCISES- Q11 TO Q20
EXERCISES- Q21 TO Q30
EXERCISES- Q31 TO Q40
EXERCISES- Q41 TO Q50
EXERCISES- Q51 TO Q58 (2-Extra Questions)
CHAPTER- 11 - Gravitation
EXERCISES -Q 31 TO 39
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
CHAPTER- 16 - Sound Waves
CHAPTER- 15 - Wave Motion and Waves on a String
EXERCISES - Q-1 TO Q-10
EXERCISES - Q-11 TO Q-20
EXERCISES - Q-21 TO Q-30
EXERCISES - Q-31 TO Q-40
EXERCISES - Q-41 TO Q-50
CHAPTER- 14 - Fluid Mechanics
CHAPTER- 13 - Fluid Mechanics
CHAPTER- 12 - Simple Harmonic Motion
EXERCISES- Q11 TO Q20
EXERCISES- Q21 TO Q30
EXERCISES- Q31 TO Q40
EXERCISES- Q41 TO Q50
EXERCISES- Q51 TO Q58 (2-Extra Questions)
CHAPTER- 11 - Gravitation
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
CHAPTER- 8 - Work and Energy
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CHAPTER- 7 - Circular Motion
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CHAPTER- 6 - Friction
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CHAPTER- 5 - Newton's Laws of Motion
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-------------------------------------------------------------------------------
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CHAPTER- 4 - The Forces
The Forces-
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CHAPTER- 3 - Kinematics - Rest and Motion
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CHAPTER- 2 - "Vector related Problems"
CHAPTER- 2 - "Vector related Problems"
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