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SOUND WAVES
EXERCISES- Q-31 to Q-40
31. Two speakers S₁ and S₂, driven by the same amplifier, are placed at y = 1.0 m and y = -1.0 m (figure 16-E4). The speakers vibrate in phase at 600 Hz. A man stands at a point on the X-axis at a very large distance from the origin and starts moving parallel to Y-axis. The speed of sound in air is 330 m/s.
(a) At what angle θ will the intensity of sound drop to a minimum for the first time?
(b) At what angle will he hear a maximum of sound intensity for the first time?
(c) If he continues to walk along the line, how many more maxima can he hear?
Figure for Q-31
ANSWER: The frequency of sound ν = 600 Hz, V = 330 m/s, Hence the wavelength 𝜆 =V/ν =330/600 m =0.55 m.
Diagram for Q-31
(a) Let the intensity of sound at Q (PQ =y) be the minimum. The path difference at Q, Δx =QS₂-QS₁
Now QS₂²-QS₁² =SS₂²+QS²-S₁R²-QR²
→(QS₂-QS₁)(QS₂+QS₁) =QS²-QR²
→Δx*2d = (y+1)²-(y-1)²=2y+2y =4y
{Since P is far away from the source, QS₂≈QS₁≈QO≈d}
→Δx=2y/d
for the destructive inerference Δx =(2n+1)λ/2
For the first minimum intensity put n=0, i.e. Δx =𝜆/2
Equating,
2y/d = 𝜆/2
→y/d = 𝜆/4
→θ = 𝜆/4, {Since for small θ = L/R}
→θ = 0.55/4 rad =0.1375 rad =7.9°
(b) For the maximum intensity Δx = n𝜆
→2y/d = n𝜆
For the first maximum intensity, n = 1
→2y/d =𝜆
→y/d = θ = 𝜆/2 =0.55/2 rad =0.275 rad =15.8°
(c) The next maxima can be found out by putting n = 2, 3, 4.... in 2y/d = n𝜆 i.e. y/d =n𝜆/2
→θ =y/d = 𝜆, 1.5𝜆, 2𝜆, 2.5𝜆, 3𝜆....
→θ = 0.55, 0.825, 1.1, 1.375, 1.65..... rad
→θ = 31.5°, 47.2°,63°, 79°, 94.5°
Here θ > 90° is not possible. Hence four more maxima can be heard.
Note: In this solution, the assumption θ = L/R is true only for small values of θ. Hence the results for higher values of θ are very rough approximations.
32. Three sources of sound S₁, S₂ and S₃ of equal intensity are placed in a straight line with S₁S₂ = S₂S₃ (figure 16-E5). At a point P, far away from the sources, the wave coming from S₂ is 120° ahead of that from S₁. Also, the wave coming from S₃ is 120° ahead of that from S₂. What would be the resultant intensity of sound at P?
Figure for Q-32
ANSWER: Since the intensity is the same hence the pressure amplitude will also be the same for each source. Let it be p₀. We can write the wave equation for S₁ at P as
p₁ = p₀ sin(kx-⍵t) =p₀ sinΦ, where Φ =kx-⍵t.
Given δ = 120° =2π/3.
Since the point P is far away we can write the wave equation at P for S₂,
p₂ = p₀ sin(Φ+2π/3), and for S₃
p₃ = p₀ sin(Φ+2π/3+2π/3) =p₀ sin(Φ+4π/3)
→p₃ = -p₀ sin(Φ+π/3)
The resultant pressure at P,
p=p₁+p₂+p₃ =p₀[sinΦ+sin(Φ+2π/3)-sin(Φ+π/3)]
=p₀[sinΦ+2cos(Φ+π/2)sin(π/3)]
=p₀[sinΦ-sinΦ] =0
Since the intensity is proportional to the square of p₀, hence the resultant intensity of sound at P =zero.
33. Two coherent narrow slits emitting sound of wavelength 𝜆 in the same phase are placed parallel to each other at a small separation of 2𝜆. The sound is detected by moving a detector on the screen Σ at a distance D(>>𝜆) from the slit S₁ as shown in figure (16-E6). Find the distance x such that the intensity at P is equal to the intensity at O.
Figure for Q-33
ANSWER: The path difference at O = OS₁-OS₂. =S₁S₂ =2𝜆. Since it is an integral multiple of the wavelength 𝜆, the interference here is constructive. Let the intensity at O = I. It is given that D >>𝜆, hence at the next constructive interference at P the intensity will remain the same. Thus we have to find the path difference at P which is a multiple of 𝜆.
S₁P²-S₂P² ={D²+x²} -{(D-2𝜆)²+x²}
→(S₁P-S₂P)(S₁P+S₂P) =4D𝜆+4𝜆²
→(S₁P-S₂P)(S₁P+S₂P) = 4D𝜆
{Since D>>𝜆, neglecting 𝜆² term and S₁P≈S₂P}
→(S₁P-S₂P){2√(D²+x²)} = 4D𝜆
→S₁P-S₂P =4D𝜆/{2√(D²+x²)}, this is the path difference for P which should be nλ for constructive interference.
→4D𝜆/{2√(D²+x²)} = n𝜆
→√(D²+x²) =2D/n
→D²+x² = 4D²/n²
→x² =4D²/n² - D²
for n = 1,
x² = 3D²
→x = √3D
There will be other value also if we put n = 2, which is x = 0, this is the point O.
34. Figure (16-E7) shows two coherent sources S₁ and S₂ which emit sound of wavelength 𝜆 in phase. The separation between the sources is 3𝜆. A circular wire of large radius is placed in such a way that S₁S₂ lies in its plane and the middle point of S₁S₂ is at the center of the wire. Find the angular position theta on the wire for which constructive interference takes place.
Figure for Q-34
ANSWER: Let the radius of the circle =R. P is a point on the wire when the radius joining it makes an angle θ with S₁S₂. The interference at P will be constructive when the path difference PS₁ - PS₂ is an integral multiple of the wavelength 𝜆 (i.e. the phase difference between both waves at P is 2nπ).
Here S₁S₂ = 3𝜆
→S₁O = S₂O = 3𝜆/2
Diagram for Q-34
Draw a perpendicular PQ on the diameter.
PQ = R sinθ, and OQ = R cosθ
Hence S₂Q = OQ - OS₂ =R cosθ - 3𝜆/2, and
S₁Q = 3𝜆 + R cosθ - 3𝜆/2 =R cosθ + 3𝜆/2.
Now PS₁² = PQ² + S₁Q²
and PS₂² = PQ² + S₂Q²
So, PS₁² - PS₂² = S₁Q² - S₂Q²
→(PS₁-PS₂)(PS₁+PS₂)=(Rcosθ+3𝜆/2)²-(Rcosθ-3𝜆/2)²
→(PS₁-PS₂)(2R) =6𝜆R cosθ
{Since R is large assuming PS₁≈PS₂}
→PS₁-PS₂ =3𝜆 cosθ
For the constructive interference, this should be equal to an integral multiple of the wavelength. Hence
3𝜆 cosθ = n𝜆
→cosθ = n/3
For n = 0,
cosθ=0 → θ = 90°
for n = 1, cosθ = 1/3
→θ = 70.5°
For n = 2, cosθ = 2/3 =0.667
→θ = 48.2°
for n = 3, cosθ = 1
→θ = 0°
Hence the constructive interference will be for θ = 0°, 48.2°, 70.5°, 90° and similar points in other quadrants.
35. Two sources of sound S₁ and S₂ vibrate at the same frequency and are in phase (figure 16-E8). The intensity of sound detected at a point P as shown in the figure is I₀.
(a) If θ equals 45°, what will be the intensity of sound detected at this point if one of the sources are switched off?
(b) What will be the answer to the previous part if θ = 60°?
Figure for Q-35
ANSWER: (a) Let the intensity at P due to one source = I and the pressure amplitude = p₀.
When both sources are active the amplitude of interfering wave = 2p₀ {Since both are in phase}. The intensity is proportional to the square of the pressure amplitude, hence
I/I₀ = p₀²/(2p₀)² = 1/4
→I = I₀/4
(b) Since the above result is independent of θ, so the intensity with only one source active will be the same even if θ = 60°. i.e. I = I₀/4
36. Find the fundamental, first overtone and second overtone frequencies of an open organ pipe of length 20 cm. The speed of sound in air is 340 m/s.
ANSWER: 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
37. A closed organ pipe can vibrate at a minimum frequency of 500 Hz. Find the length of the tube. The speed of sound in air = 340 m/s.
ANSWER: The minimum frequency of the vibration is the fundamental frequency of the vibration. In a closed organ pipe, the fundamental frequency of vibration is given as,
ν = V/4L, where V = speed of the sound in air = 340 m/s and ν = 500 Hz. L = length of the pipe =?
Thus, L =V/4ν =340/(4*500) m =340/2000 m
→L = 0.17 m = 17 cm.
38. In a standing wave pattern in a vibrating air column, nodes are formed at a distance of 4.0 cm. If the speed of sound in air is 328 m/s, what is the frequency of the source?
ANSWER: 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
𝜆/2 = 0.04 m
→𝜆 = 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/𝜆
→ν = 328/0.08 Hz =32800/8 Hz
→ν = 4100 Hz = 4.10 kHz.
39. The separation between a node and the next antinode in a vibrating air column is 25 cm. If the speed of sound in air is 340 m/s, find the frequency of vibration of the air column.
ANSWER: 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
40. A cylindrical metal tube has a length of 50 cm and is open at both ends. Find the frequencies between 1000 Hz and 2000 Hz at which the air column in the tube can resonate. The speed of sound in air is 340 m/s.
ANSWER: 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.
31. Two speakers S₁ and S₂, driven by the same amplifier, are placed at y = 1.0 m and y = -1.0 m (figure 16-E4). The speakers vibrate in phase at 600 Hz. A man stands at a point on the X-axis at a very large distance from the origin and starts moving parallel to Y-axis. The speed of sound in air is 330 m/s.
(a) At what angle θ will the intensity of sound drop to a minimum for the first time?
(b) At what angle will he hear a maximum of sound intensity for the first time?
(c) If he continues to walk along the line, how many more maxima can he hear?
ANSWER: The frequency of sound ν = 600 Hz, V = 330 m/s, Hence the wavelength 𝜆 =V/ν =330/600 m =0.55 m.
(a) Let the intensity of sound at Q (PQ =y) be the minimum. The path difference at Q, Δx =QS₂-QS₁
Now QS₂²-QS₁² =SS₂²+QS²-S₁R²-QR²
→(QS₂-QS₁)(QS₂+QS₁) =QS²-QR²
→Δx*2d = (y+1)²-(y-1)²=2y+2y =4y
{Since P is far away from the source, QS₂≈QS₁≈QO≈d}
→Δx=2y/d
for the destructive inerference Δx =(2n+1)λ/2
For the first minimum intensity put n=0, i.e. Δx =𝜆/2
Equating,
2y/d = 𝜆/2
→y/d = 𝜆/4
→θ = 𝜆/4, {Since for small θ = L/R}
→θ = 0.55/4 rad =0.1375 rad =7.9°
(b) For the maximum intensity Δx = n𝜆
→2y/d = n𝜆
For the first maximum intensity, n = 1
→2y/d =𝜆
→y/d = θ = 𝜆/2 =0.55/2 rad =0.275 rad =15.8°
(c) The next maxima can be found out by putting n = 2, 3, 4.... in 2y/d = n𝜆 i.e. y/d =n𝜆/2
→θ =y/d = 𝜆, 1.5𝜆, 2𝜆, 2.5𝜆, 3𝜆....
→θ = 0.55, 0.825, 1.1, 1.375, 1.65..... rad
→θ = 31.5°, 47.2°,63°, 79°, 94.5°
Here θ > 90° is not possible. Hence four more maxima can be heard.
Note: In this solution, the assumption θ = L/R is true only for small values of θ. Hence the results for higher values of θ are very rough approximations.
32. Three sources of sound S₁, S₂ and S₃ of equal intensity are placed in a straight line with S₁S₂ = S₂S₃ (figure 16-E5). At a point P, far away from the sources, the wave coming from S₂ is 120° ahead of that from S₁. Also, the wave coming from S₃ is 120° ahead of that from S₂. What would be the resultant intensity of sound at P?
ANSWER: Since the intensity is the same hence the pressure amplitude will also be the same for each source. Let it be p₀. We can write the wave equation for S₁ at P as
p₁ = p₀ sin(kx-⍵t) =p₀ sinΦ, where Φ =kx-⍵t.
Given δ = 120° =2π/3.
Since the point P is far away we can write the wave equation at P for S₂,
p₂ = p₀ sin(Φ+2π/3), and for S₃
p₃ = p₀ sin(Φ+2π/3+2π/3) =p₀ sin(Φ+4π/3)
→p₃ = -p₀ sin(Φ+π/3)
The resultant pressure at P,
p=p₁+p₂+p₃ =p₀[sinΦ+sin(Φ+2π/3)-sin(Φ+π/3)]
=p₀[sinΦ+2cos(Φ+π/2)sin(π/3)]
=p₀[sinΦ-sinΦ] =0
Since the intensity is proportional to the square of p₀, hence the resultant intensity of sound at P =zero.
33. Two coherent narrow slits emitting sound of wavelength 𝜆 in the same phase are placed parallel to each other at a small separation of 2𝜆. The sound is detected by moving a detector on the screen Σ at a distance D(>>𝜆) from the slit S₁ as shown in figure (16-E6). Find the distance x such that the intensity at P is equal to the intensity at O.
ANSWER: The path difference at O = OS₁-OS₂. =S₁S₂ =2𝜆. Since it is an integral multiple of the wavelength 𝜆, the interference here is constructive. Let the intensity at O = I. It is given that D >>𝜆, hence at the next constructive interference at P the intensity will remain the same. Thus we have to find the path difference at P which is a multiple of 𝜆.
S₁P²-S₂P² ={D²+x²} -{(D-2𝜆)²+x²}
→(S₁P-S₂P)(S₁P+S₂P) =4D𝜆+4𝜆²
→(S₁P-S₂P)(S₁P+S₂P) = 4D𝜆
{Since D>>𝜆, neglecting 𝜆² term and S₁P≈S₂P}
→(S₁P-S₂P){2√(D²+x²)} = 4D𝜆
→S₁P-S₂P =4D𝜆/{2√(D²+x²)}, this is the path difference for P which should be nλ for constructive interference.
→4D𝜆/{2√(D²+x²)} = n𝜆
→√(D²+x²) =2D/n
→D²+x² = 4D²/n²
→x² =4D²/n² - D²
for n = 1,
x² = 3D²
→x = √3D
There will be other value also if we put n = 2, which is x = 0, this is the point O.
34. Figure (16-E7) shows two coherent sources S₁ and S₂ which emit sound of wavelength 𝜆 in phase. The separation between the sources is 3𝜆. A circular wire of large radius is placed in such a way that S₁S₂ lies in its plane and the middle point of S₁S₂ is at the center of the wire. Find the angular position theta on the wire for which constructive interference takes place.
ANSWER: Let the radius of the circle =R. P is a point on the wire when the radius joining it makes an angle θ with S₁S₂. The interference at P will be constructive when the path difference PS₁ - PS₂ is an integral multiple of the wavelength 𝜆 (i.e. the phase difference between both waves at P is 2nπ).
Here S₁S₂ = 3𝜆
→S₁O = S₂O = 3𝜆/2
Draw a perpendicular PQ on the diameter.
PQ = R sinθ, and OQ = R cosθ
Hence S₂Q = OQ - OS₂ =R cosθ - 3𝜆/2, and
S₁Q = 3𝜆 + R cosθ - 3𝜆/2 =R cosθ + 3𝜆/2.
Now PS₁² = PQ² + S₁Q²
and PS₂² = PQ² + S₂Q²
So, PS₁² - PS₂² = S₁Q² - S₂Q²
→(PS₁-PS₂)(PS₁+PS₂)=(Rcosθ+3𝜆/2)²-(Rcosθ-3𝜆/2)²
→(PS₁-PS₂)(2R) =6𝜆R cosθ
{Since R is large assuming PS₁≈PS₂}
→PS₁-PS₂ =3𝜆 cosθ
For the constructive interference, this should be equal to an integral multiple of the wavelength. Hence
3𝜆 cosθ = n𝜆
→cosθ = n/3
For n = 0,
cosθ=0 → θ = 90°
for n = 1, cosθ = 1/3
→θ = 70.5°
For n = 2, cosθ = 2/3 =0.667
→θ = 48.2°
for n = 3, cosθ = 1
→θ = 0°
Hence the constructive interference will be for θ = 0°, 48.2°, 70.5°, 90° and similar points in other quadrants.
35. Two sources of sound S₁ and S₂ vibrate at the same frequency and are in phase (figure 16-E8). The intensity of sound detected at a point P as shown in the figure is I₀.
(a) If θ equals 45°, what will be the intensity of sound detected at this point if one of the sources are switched off?
(b) What will be the answer to the previous part if θ = 60°?
ANSWER: (a) Let the intensity at P due to one source = I and the pressure amplitude = p₀.
When both sources are active the amplitude of interfering wave = 2p₀ {Since both are in phase}. The intensity is proportional to the square of the pressure amplitude, hence
I/I₀ = p₀²/(2p₀)² = 1/4
→I = I₀/4
(b) Since the above result is independent of θ, so the intensity with only one source active will be the same even if θ = 60°. i.e. I = I₀/4
36. Find the fundamental, first overtone and second overtone frequencies of an open organ pipe of length 20 cm. The speed of sound in air is 340 m/s.
ANSWER: 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
37. A closed organ pipe can vibrate at a minimum frequency of 500 Hz. Find the length of the tube. The speed of sound in air = 340 m/s.
ANSWER: The minimum frequency of the vibration is the fundamental frequency of the vibration. In a closed organ pipe, the fundamental frequency of vibration is given as,
ν = V/4L, where V = speed of the sound in air = 340 m/s and ν = 500 Hz. L = length of the pipe =?
Thus, L =V/4ν =340/(4*500) m =340/2000 m
→L = 0.17 m = 17 cm.
38. In a standing wave pattern in a vibrating air column, nodes are formed at a distance of 4.0 cm. If the speed of sound in air is 328 m/s, what is the frequency of the source?
ANSWER: 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
𝜆/2 = 0.04 m
→𝜆 = 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/𝜆
→ν = 328/0.08 Hz =32800/8 Hz
→ν = 4100 Hz = 4.10 kHz.
39. The separation between a node and the next antinode in a vibrating air column is 25 cm. If the speed of sound in air is 340 m/s, find the frequency of vibration of the air column.
ANSWER: 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
40. A cylindrical metal tube has a length of 50 cm and is open at both ends. Find the frequencies between 1000 Hz and 2000 Hz at which the air column in the tube can resonate. The speed of sound in air is 340 m/s.
ANSWER: 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.
(a) At what angle θ will the intensity of sound drop to a minimum for the first time?
(b) At what angle will he hear a maximum of sound intensity for the first time?
(c) If he continues to walk along the line, how many more maxima can he hear?
 |
| Figure for Q-31 |
ANSWER: The frequency of sound ν = 600 Hz, V = 330 m/s, Hence the wavelength 𝜆 =V/ν =330/600 m =0.55 m.
|
| Diagram for Q-31 |
(a) Let the intensity of sound at Q (PQ =y) be the minimum. The path difference at Q, Δx =QS₂-QS₁
Now QS₂²-QS₁² =SS₂²+QS²-S₁R²-QR²
→(QS₂-QS₁)(QS₂+QS₁) =QS²-QR²
→Δx*2d = (y+1)²-(y-1)²=2y+2y =4y
{Since P is far away from the source, QS₂≈QS₁≈QO≈d}
→Δx=2y/d
for the destructive inerference Δx =(2n+1)λ/2
For the first minimum intensity put n=0, i.e. Δx =𝜆/2
Equating,
2y/d = 𝜆/2
→y/d = 𝜆/4
→θ = 𝜆/4, {Since for small θ = L/R}
→θ = 0.55/4 rad =0.1375 rad =7.9°
(b) For the maximum intensity Δx = n𝜆
→2y/d = n𝜆
For the first maximum intensity, n = 1
→2y/d =𝜆
→y/d = θ = 𝜆/2 =0.55/2 rad =0.275 rad =15.8°
(c) The next maxima can be found out by putting n = 2, 3, 4.... in 2y/d = n𝜆 i.e. y/d =n𝜆/2
→θ =y/d = 𝜆, 1.5𝜆, 2𝜆, 2.5𝜆, 3𝜆....
→θ = 0.55, 0.825, 1.1, 1.375, 1.65..... rad
→θ = 31.5°, 47.2°,63°, 79°, 94.5°
Here θ > 90° is not possible. Hence four more maxima can be heard.
Note: In this solution, the assumption θ = L/R is true only for small values of θ. Hence the results for higher values of θ are very rough approximations.
32. Three sources of sound S₁, S₂ and S₃ of equal intensity are placed in a straight line with S₁S₂ = S₂S₃ (figure 16-E5). At a point P, far away from the sources, the wave coming from S₂ is 120° ahead of that from S₁. Also, the wave coming from S₃ is 120° ahead of that from S₂. What would be the resultant intensity of sound at P?
 |
| Figure for Q-32 |
ANSWER: Since the intensity is the same hence the pressure amplitude will also be the same for each source. Let it be p₀. We can write the wave equation for S₁ at P as
p₁ = p₀ sin(kx-⍵t) =p₀ sinΦ, where Φ =kx-⍵t.
Given δ = 120° =2π/3.
Since the point P is far away we can write the wave equation at P for S₂,
p₂ = p₀ sin(Φ+2π/3), and for S₃
p₃ = p₀ sin(Φ+2π/3+2π/3) =p₀ sin(Φ+4π/3)
→p₃ = -p₀ sin(Φ+π/3)
The resultant pressure at P,
p=p₁+p₂+p₃ =p₀[sinΦ+sin(Φ+2π/3)-sin(Φ+π/3)]
=p₀[sinΦ+2cos(Φ+π/2)sin(π/3)]
=p₀[sinΦ-sinΦ] =0
Since the intensity is proportional to the square of p₀, hence the resultant intensity of sound at P =zero.
33. Two coherent narrow slits emitting sound of wavelength 𝜆 in the same phase are placed parallel to each other at a small separation of 2𝜆. The sound is detected by moving a detector on the screen Σ at a distance D(>>𝜆) from the slit S₁ as shown in figure (16-E6). Find the distance x such that the intensity at P is equal to the intensity at O.
 |
| Figure for Q-33 |
ANSWER: The path difference at O = OS₁-OS₂. =S₁S₂ =2𝜆. Since it is an integral multiple of the wavelength 𝜆, the interference here is constructive. Let the intensity at O = I. It is given that D >>𝜆, hence at the next constructive interference at P the intensity will remain the same. Thus we have to find the path difference at P which is a multiple of 𝜆.
S₁P²-S₂P² ={D²+x²} -{(D-2𝜆)²+x²}
→(S₁P-S₂P)(S₁P+S₂P) =4D𝜆+4𝜆²
→(S₁P-S₂P)(S₁P+S₂P) = 4D𝜆
{Since D>>𝜆, neglecting 𝜆² term and S₁P≈S₂P}
→(S₁P-S₂P){2√(D²+x²)} = 4D𝜆
→S₁P-S₂P =4D𝜆/{2√(D²+x²)}, this is the path difference for P which should be nλ for constructive interference.
→4D𝜆/{2√(D²+x²)} = n𝜆
→√(D²+x²) =2D/n
→D²+x² = 4D²/n²
→x² =4D²/n² - D²
for n = 1,
x² = 3D²
→x = √3D
There will be other value also if we put n = 2, which is x = 0, this is the point O.
34. Figure (16-E7) shows two coherent sources S₁ and S₂ which emit sound of wavelength 𝜆 in phase. The separation between the sources is 3𝜆. A circular wire of large radius is placed in such a way that S₁S₂ lies in its plane and the middle point of S₁S₂ is at the center of the wire. Find the angular position theta on the wire for which constructive interference takes place.
|
| Figure for Q-34 |
ANSWER: Let the radius of the circle =R. P is a point on the wire when the radius joining it makes an angle θ with S₁S₂. The interference at P will be constructive when the path difference PS₁ - PS₂ is an integral multiple of the wavelength 𝜆 (i.e. the phase difference between both waves at P is 2nπ).
Here S₁S₂ = 3𝜆
→S₁O = S₂O = 3𝜆/2
|
| Diagram for Q-34 |
Draw a perpendicular PQ on the diameter.
PQ = R sinθ, and OQ = R cosθ
Hence S₂Q = OQ - OS₂ =R cosθ - 3𝜆/2, and
S₁Q = 3𝜆 + R cosθ - 3𝜆/2 =R cosθ + 3𝜆/2.
Now PS₁² = PQ² + S₁Q²
and PS₂² = PQ² + S₂Q²
So, PS₁² - PS₂² = S₁Q² - S₂Q²
→(PS₁-PS₂)(PS₁+PS₂)=(Rcosθ+3𝜆/2)²-(Rcosθ-3𝜆/2)²
→(PS₁-PS₂)(2R) =6𝜆R cosθ
{Since R is large assuming PS₁≈PS₂}
→PS₁-PS₂ =3𝜆 cosθ
For the constructive interference, this should be equal to an integral multiple of the wavelength. Hence
3𝜆 cosθ = n𝜆
→cosθ = n/3
For n = 0,
cosθ=0 → θ = 90°
for n = 1, cosθ = 1/3
→θ = 70.5°
For n = 2, cosθ = 2/3 =0.667
→θ = 48.2°
for n = 3, cosθ = 1
→θ = 0°
Hence the constructive interference will be for θ = 0°, 48.2°, 70.5°, 90° and similar points in other quadrants.
35. Two sources of sound S₁ and S₂ vibrate at the same frequency and are in phase (figure 16-E8). The intensity of sound detected at a point P as shown in the figure is I₀.
(a) If θ equals 45°, what will be the intensity of sound detected at this point if one of the sources are switched off?
(b) What will be the answer to the previous part if θ = 60°?
 |
| Figure for Q-35 |
ANSWER: (a) Let the intensity at P due to one source = I and the pressure amplitude = p₀.
When both sources are active the amplitude of interfering wave = 2p₀ {Since both are in phase}. The intensity is proportional to the square of the pressure amplitude, hence
I/I₀ = p₀²/(2p₀)² = 1/4
→I = I₀/4
(b) Since the above result is independent of θ, so the intensity with only one source active will be the same even if θ = 60°. i.e. I = I₀/4
36. Find the fundamental, first overtone and second overtone frequencies of an open organ pipe of length 20 cm. The speed of sound in air is 340 m/s.
ANSWER: 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
37. A closed organ pipe can vibrate at a minimum frequency of 500 Hz. Find the length of the tube. The speed of sound in air = 340 m/s.
ANSWER: The minimum frequency of the vibration is the fundamental frequency of the vibration. In a closed organ pipe, the fundamental frequency of vibration is given as,
ν = V/4L, where V = speed of the sound in air = 340 m/s and ν = 500 Hz. L = length of the pipe =?
Thus, L =V/4ν =340/(4*500) m =340/2000 m
→L = 0.17 m = 17 cm.
38. In a standing wave pattern in a vibrating air column, nodes are formed at a distance of 4.0 cm. If the speed of sound in air is 328 m/s, what is the frequency of the source?
ANSWER: 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
𝜆/2 = 0.04 m
→𝜆 = 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/𝜆
→ν = 328/0.08 Hz =32800/8 Hz
→ν = 4100 Hz = 4.10 kHz.
39. The separation between a node and the next antinode in a vibrating air column is 25 cm. If the speed of sound in air is 340 m/s, find the frequency of vibration of the air column.
ANSWER: 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
40. A cylindrical metal tube has a length of 50 cm and is open at both ends. Find the frequencies between 1000 Hz and 2000 Hz at which the air column in the tube can resonate. The speed of sound in air is 340 m/s.
ANSWER: 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.
===<<<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
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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
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EXERCISES - Q-11 TO Q-20
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CHAPTER- 14 - Fluid Mechanics
CHAPTER- 13 - Fluid Mechanics
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EXERCISES- Q11 TO Q20
EXERCISES- Q21 TO Q30
EXERCISES- Q31 TO Q40
EXERCISES- Q41 TO Q50
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CHAPTER- 11 - Gravitation
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CHAPTER- 5 - Newton's Laws of Motion
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CHAPTER- 4 - 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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