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SOUND WAVES
EXERCISES- Q-11 to Q-20
11. Calculate the speed of sound in oxygen from the following data. The mass of 22.4 liters of oxygen at STP (T = 273 K and p = 1.0x10⁵ N/m²) is 32 g, the molar heat capacity of oxygen at constant volume is Cᵥ = 2.5 R and that at constant pressure Cₚ = 3.5 R.
ANSWER: The speed of sound in a gas is given as
V =√{𝜸P/ρ}
Here 𝜸 = Cₚ/Cᵥ =3.5 /2.5 =1.4
P = 1.0 x 10⁵ N/m²
ρ = mass/volume = (32/1000 kg)/(22.4/1000 m³)
=32/22.4 kg/m³
Hence the speed of sound in oxygen
V =√{1.4*1.0x10⁵ *22.4/32} m/s
=√(98000) m/s
=313 m/s
12. The speed of sound as measured by a student in the laboratory on a winter day is 340 m/s when the room temperature is 17 °C. What speed will be measured by another student repeating the experiment on a day when the room temperature is 32 °C?
ANSWER: The speed of sound is proportional to the square root of the temperature, hence
V/V' = √T/√T'
→V' =V√T'/√T
Here speed at 17 °C, V = 340 m/s,
T = 273 + 17 K =290 K
T' = 273 + 32 K =305 K
Speed at 32 °C, V' = 340*√305/√290 m/s
→V' = 340*1.026 m/s =349 m/s
13. At what temperature will the speed of sound be double of its value at 0 °C?
ANSWER: Let the speed of sound be double at t °C than at 0 °C.
T =0 °C =273 K
T' = t °C =273 + t K
speed of sound at 0 °C = V, speed of spond at t °C = V' =2V,
V' = V√T'/√T
→2V = V√(T'/T)
→2 = √(T'/T)
→T'/T =4
→T' =4T =4*273 =1092
→273+t =1092
→t = 1092-273 °C =819 °C
14. The absolute temperature of air in a region linearly increases from T₁ to T₂ in a space of width d. Find the time taken by a sound wave to go through the region in terms of T₁, T₂, d and the speed v of sound at 273 K. Evaluate this time for T₁ = 280 K, T₂ = 310 K, d = 33 m and v = 330 m/s.
ANSWER: The temperature of the air at a distance x from the T₁ end,
Tₓ = T₁ + (T₂ - T₁)x/d
Diagram for Q-14
The speed of sound at x will be
Vₓ = V√Tₓ/√T
where V is the speed of sound at 0 °C.
→Vₓ =V√{T₁+(T₂ - T₁)x/d}/√T
=V√[{T₁d+(T₂ - T₁)x}/Td]
Consider a small distance dx. The small time dt taken by the sound waves to travel the distance dx is given by
dt = dx/Vₓ
→dt = dx/[V√[{T₁d+(T₂ - T₁)x}/Td]
→dt = dx/[V√[{T₁/T+(T₂ - T₁)x/Td]]
Integrating dt for x = 0 to x = d,
t =∫dt =∫dx/[V√[T₁/T+(T₂ - T₁)x/Td]
=2[V√[T₁/T+(T₂ - T₁)x/Td]/[V²(T₂-T₁)/Td]
=2Td√[T₁/T+(T₂ - T₁)x/Td]/V(T₂-T₁)
=2√[Td{T₁d+(T₂ - T₁)x}]/V(T₂-T₁)
=2{√T√d/V}[√{T₁d+(T₂ - T₁)x}/(T₂-T₁)]
putting the integration values
=2{√T√d/V(T₂-T₁)}[√{T₁d+(T₂ - T₁)d} -√(T₁d)]
=2{√T√d/V(T₂-T₁)}[√(T₂d) - √(T₁d)]
=2{√T√d/V(T₂-T₁)}[(√T₂ - √T₁)√d]
=2√T*d(√T₂ - √T₁)/V(T₂ - T₁)
=2√T*d(√T₂ - √T₁)/V{(√T₂)² - (√T₁)²}
=2d√T/V(√T₂ + √T₁)
(Putting value T = 273 K)
=2d√273/V(√T₂ + √T₁)
[NOTE: Here the integration is based on
∫dx/√(a+bx) =2√(a+bx)/b]
For T₁ = 280 K, T₂ = 310 K, d = 33 m and V =330 m/s the time t =2*33*√273/330(√310 +√280) s
=3.30/(17.61+16.73) s
=3.30/34.34 s
=0.096 s
=96 ms
15. Find the change in the volume of 1.0-liter kerosene when it is subjected to an extra pressure of 2.0 x 10⁵ N/m² from the following data. Density of kerosene = 800 kg/m³ and speed of sound in kerosene = 1330 m/s.
ANSWER: The speed of sound V =√(B/ρ)
where B = bulk modulus and ρ = density of the liquid.
From this B =ρV² =800*1330² N/m²
→B = 1.42x10⁹ N/m²
Given volume of Kerosene, v = 1 liter =1x10⁻³ m³
Let the change in volume with the extra pressure =Δv
So, p/(Δv/v) = B, where p = pressure.
→Δv/v = p/B
→Δv = pv/B =2.0x10⁵*1x10⁻³/1.42x10⁹ m³
=1.40x10⁻⁷ m³
=1.40x10⁻⁷*10⁶ cm³
=1.40x0.10 cm³
=0.14 cm³
16. Calculate the bulk modulus of air from the following data about a sound wave of wavelength 35 cm traveling in air. The pressure at a point varies between (1 x 10⁵ 土 14) Pa and the particles of the air vibrate in a simple harmonic motion of amplitude 5.5 x 10⁻⁶ m.
ANSWER: The pressure amplitude p₀ =14 Pa
The displacement amplitude s₀ = 5.5x10⁻⁶ m
Wavelength, λ = 35 cm =0.35 m
Wave number k =2π/λ
The relation between p₀, s₀, k and B is given by
p₀ = Bks₀ =2πBs₀/λ, {Since k = 2π/λ}
→B = p₀λ/2πs₀ =14*0.35/(2π*5.5x10⁻⁶) N/m²
→B = 0.14x10⁶ N/m²
→B = 1.4x10⁵ N/m²
17. A source of sound operates 2.0 kHz, 20 W emitting sound uniformly in all directions. The speed of sound in air is 340 m/s and the density of air is 1.2 kg /m³.
(a) What is the intensity at a distance of 6.0 m from the source?
(b) What will be the pressure amplitude at this point?
(c) What will be the displacement amplitude at this point?
ANSWER: (a) The intensity of sound is the energy passing through a unit area in unit time. Here 20 W i.e. 20 J of energy is passing per second through a spherical surface having radius 6.0 m. Area of this surface =4π*6² m² = 144π m².
Hence the intensity of sound at 6.0 m from the source, I = 20/144π W/m²
= 0.044 W/m²
= 44 mW/m²
(b) The relation between the intensity of sound and the pressure amplitude is given as
I = p₀²/2ρV
→p₀² = 2ρVI
Given that,
V = 340 m/s, ρ = 1.2 kg/m³, from (a) above I =0.044 W/m²
Hence p₀² =2*1.2*340*0.044
→p₀² = 36
→p₀ = 6.0 Pa
(c) Given frequency 𝜈 = 2.0 kHz =2000 Hz
The intensity is also given as
I =p₀²V/2B and the relation between p₀ and s₀ (displacement amplitude) is given as p₀ = B⍵s₀/V i.e. B = p₀V/⍵s₀. Eliminating B, we have
I = p₀²V⍵s₀/2p₀V =p₀⍵s₀/2 =p₀(2π𝜈)s₀/2
→0.044 = 6*(2π*2000)s₀/2
→s₀ = 0.044/12000π m =1.2x10⁻⁶ m
18. The intensity of sound from a point source is 1.0 x10⁻⁸ W/m² at a distance of 5.0 m from the source. What will be the intensity at a distance of 25 m from the source?
Diagram for Q-18
ANSWER: Consider a square of side DE perpendicular to the line joining the source and a point 25 m from it. Join the vertices of the square to the point source thus forming a pyramid with vertex at the source A. Now consider the section at a distance 5 m from the vertex and parallel to the first square. It will be another square with a side BC. In the similar triangles ABC and ADE,
BC/DE = 5/25 =1/5
→BC = DE/5
So the side of the square at 5 m distance will be one-fifth of the square at 25 m, Hence the area of the square at 5 m will be one-twenty-fifth of the area of the square at 25 m. It is clear from the figure that the same amount of energy (say U) is passing through both areas. If the area of the square at 25 m =A then the area of the square at 5 m =A/25.
The Intensity at 25 m, I = U/A and the intensity at 5 m, I' = U/(A/25) =25(U/A) =25*I
→I =I'/25.
Given that I' = 1.0 x10⁻⁸ W/m²
Hence I = 1.0 x10⁻⁸/25 W/m² = 0.04x10⁻⁸ W/m² =4.0x10⁻¹⁰ W/m²
(It is clear from this diagram that the intensity at a point is inversely proportional to the square of the distance from a point source)
19. The sound level at a point 5.0 m away from a point source is 40 dB. What will be the level at a point 50 m away from the source?
ANSWER: The sound level is given as
ß = 10 log₁₀ [I/I₀]
Given that ß₅ = 40 dB
If the intensity at 5.0 m = I₁ and the intensity at 50 m = I₂, then
The sound level at 5.0 m ß₅ = 10 log₁₀ [I₁/I₀]
The sound level at 50 m ß₅₀ =10 log₁₀ [I₂/I₀]
So, ß₅₀ - ß₅ = 10 log₁₀ [I₂/I₀] - 10 log₁₀ [I₁/I₀]
=[10 log₁₀ I₂ -10 log₁₀ I₀] - [10 log₁₀ I₁ -10 log₁₀ I₀]
= 10 log₁₀ I₂- 10 log₁₀ I₁
= 10 log₁₀ [I₂/I₁]
Since the intensity at a point is inversely proportional to the square of the distance from the point source,
I₂/I₁ = 5²/50² = 1/100
So, ß₅₀ - ß₅ = 10 log₁₀ [I₂/I₁] =10 log₁₀ [1/100]
→ ß₅₀ - ß₅ = 10[log₁₀ 1 -log₁₀ 100] = 10[0 - 2]
→ ß₅₀ - ß₅ = -20
→ ß₅₀ = ß₅ -20 = 40 - 20 = 20 dB
20. If the intensity of sound is doubled, by how many decibels does the sound level increase?
ANSWER: For the intensity I, sound level
ß = 10 log₁₀ [I/I₀]
When the intensity is 2I, sound level
ß' = 10 log₁₀ [2I/I₀]
Hence ß' - ß = 10 log₁₀ [2I/I₀]-10 log₁₀ [I/I₀]
→ß' -ß =10 log₁₀ [2I/I] = 10 log₁₀ 2 =10*0.3 =3 dB
→ß' = ß + 3 dB
So when the intensity is doubled the sound level is increased by 3 dB.
11. Calculate the speed of sound in oxygen from the following data. The mass of 22.4 liters of oxygen at STP (T = 273 K and p = 1.0x10⁵ N/m²) is 32 g, the molar heat capacity of oxygen at constant volume is Cᵥ = 2.5 R and that at constant pressure Cₚ = 3.5 R.
ANSWER: The speed of sound in a gas is given as
V =√{𝜸P/ρ}
Here 𝜸 = Cₚ/Cᵥ =3.5 /2.5 =1.4
P = 1.0 x 10⁵ N/m²
ρ = mass/volume = (32/1000 kg)/(22.4/1000 m³)
=32/22.4 kg/m³
Hence the speed of sound in oxygen
V =√{1.4*1.0x10⁵ *22.4/32} m/s
=√(98000) m/s
=313 m/s
12. The speed of sound as measured by a student in the laboratory on a winter day is 340 m/s when the room temperature is 17 °C. What speed will be measured by another student repeating the experiment on a day when the room temperature is 32 °C?
ANSWER: The speed of sound is proportional to the square root of the temperature, hence
V/V' = √T/√T'
→V' =V√T'/√T
Here speed at 17 °C, V = 340 m/s,
T = 273 + 17 K =290 K
T' = 273 + 32 K =305 K
Speed at 32 °C, V' = 340*√305/√290 m/s
→V' = 340*1.026 m/s =349 m/s
13. At what temperature will the speed of sound be double of its value at 0 °C?
ANSWER: Let the speed of sound be double at t °C than at 0 °C.
T =0 °C =273 K
T' = t °C =273 + t K
speed of sound at 0 °C = V, speed of spond at t °C = V' =2V,
V' = V√T'/√T
→2V = V√(T'/T)
→2 = √(T'/T)
→T'/T =4
→T' =4T =4*273 =1092
→273+t =1092
→t = 1092-273 °C =819 °C
14. The absolute temperature of air in a region linearly increases from T₁ to T₂ in a space of width d. Find the time taken by a sound wave to go through the region in terms of T₁, T₂, d and the speed v of sound at 273 K. Evaluate this time for T₁ = 280 K, T₂ = 310 K, d = 33 m and v = 330 m/s.
ANSWER: The temperature of the air at a distance x from the T₁ end,
Tₓ = T₁ + (T₂ - T₁)x/d
The speed of sound at x will be
Vₓ = V√Tₓ/√T
where V is the speed of sound at 0 °C.
→Vₓ =V√{T₁+(T₂ - T₁)x/d}/√T
=V√[{T₁d+(T₂ - T₁)x}/Td]
Consider a small distance dx. The small time dt taken by the sound waves to travel the distance dx is given by
dt = dx/Vₓ
→dt = dx/[V√[{T₁d+(T₂ - T₁)x}/Td]
→dt = dx/[V√[{T₁/T+(T₂ - T₁)x/Td]]
Integrating dt for x = 0 to x = d,
t =∫dt =∫dx/[V√[T₁/T+(T₂ - T₁)x/Td]
=2[V√[T₁/T+(T₂ - T₁)x/Td]/[V²(T₂-T₁)/Td]
=2Td√[T₁/T+(T₂ - T₁)x/Td]/V(T₂-T₁)
=2√[Td{T₁d+(T₂ - T₁)x}]/V(T₂-T₁)
=2{√T√d/V}[√{T₁d+(T₂ - T₁)x}/(T₂-T₁)]
putting the integration values
=2{√T√d/V(T₂-T₁)}[√{T₁d+(T₂ - T₁)d} -√(T₁d)]
=2{√T√d/V(T₂-T₁)}[√(T₂d) - √(T₁d)]
=2{√T√d/V(T₂-T₁)}[(√T₂ - √T₁)√d]
=2√T*d(√T₂ - √T₁)/V(T₂ - T₁)
=2√T*d(√T₂ - √T₁)/V{(√T₂)² - (√T₁)²}
=2d√T/V(√T₂ + √T₁)
(Putting value T = 273 K)
=2d√273/V(√T₂ + √T₁)
[NOTE: Here the integration is based on
∫dx/√(a+bx) =2√(a+bx)/b]
For T₁ = 280 K, T₂ = 310 K, d = 33 m and V =330 m/s the time t =2*33*√273/330(√310 +√280) s
=3.30/(17.61+16.73) s
=3.30/34.34 s
=0.096 s
=96 ms
15. Find the change in the volume of 1.0-liter kerosene when it is subjected to an extra pressure of 2.0 x 10⁵ N/m² from the following data. Density of kerosene = 800 kg/m³ and speed of sound in kerosene = 1330 m/s.
ANSWER: The speed of sound V =√(B/ρ)
where B = bulk modulus and ρ = density of the liquid.
From this B =ρV² =800*1330² N/m²
→B = 1.42x10⁹ N/m²
Given volume of Kerosene, v = 1 liter =1x10⁻³ m³
Let the change in volume with the extra pressure =Δv
So, p/(Δv/v) = B, where p = pressure.
→Δv/v = p/B
→Δv = pv/B =2.0x10⁵*1x10⁻³/1.42x10⁹ m³
=1.40x10⁻⁷ m³
=1.40x10⁻⁷*10⁶ cm³
=1.40x0.10 cm³
=0.14 cm³
16. Calculate the bulk modulus of air from the following data about a sound wave of wavelength 35 cm traveling in air. The pressure at a point varies between (1 x 10⁵ 土 14) Pa and the particles of the air vibrate in a simple harmonic motion of amplitude 5.5 x 10⁻⁶ m.
ANSWER: The pressure amplitude p₀ =14 Pa
The displacement amplitude s₀ = 5.5x10⁻⁶ m
Wavelength, λ = 35 cm =0.35 m
Wave number k =2π/λ
The relation between p₀, s₀, k and B is given by
p₀ = Bks₀ =2πBs₀/λ, {Since k = 2π/λ}
→B = p₀λ/2πs₀ =14*0.35/(2π*5.5x10⁻⁶) N/m²
→B = 0.14x10⁶ N/m²
→B = 1.4x10⁵ N/m²
17. A source of sound operates 2.0 kHz, 20 W emitting sound uniformly in all directions. The speed of sound in air is 340 m/s and the density of air is 1.2 kg /m³.
(a) What is the intensity at a distance of 6.0 m from the source?
(b) What will be the pressure amplitude at this point?
(c) What will be the displacement amplitude at this point?
ANSWER: (a) The intensity of sound is the energy passing through a unit area in unit time. Here 20 W i.e. 20 J of energy is passing per second through a spherical surface having radius 6.0 m. Area of this surface =4π*6² m² = 144π m².
Hence the intensity of sound at 6.0 m from the source, I = 20/144π W/m²
= 0.044 W/m²
= 44 mW/m²
(b) The relation between the intensity of sound and the pressure amplitude is given as
I = p₀²/2ρV
→p₀² = 2ρVI
Given that,
V = 340 m/s, ρ = 1.2 kg/m³, from (a) above I =0.044 W/m²
Hence p₀² =2*1.2*340*0.044
→p₀² = 36
→p₀ = 6.0 Pa
(c) Given frequency 𝜈 = 2.0 kHz =2000 Hz
The intensity is also given as
I =p₀²V/2B and the relation between p₀ and s₀ (displacement amplitude) is given as p₀ = B⍵s₀/V i.e. B = p₀V/⍵s₀. Eliminating B, we have
I = p₀²V⍵s₀/2p₀V =p₀⍵s₀/2 =p₀(2π𝜈)s₀/2
→0.044 = 6*(2π*2000)s₀/2
→s₀ = 0.044/12000π m =1.2x10⁻⁶ m
18. The intensity of sound from a point source is 1.0 x10⁻⁸ W/m² at a distance of 5.0 m from the source. What will be the intensity at a distance of 25 m from the source?
ANSWER: Consider a square of side DE perpendicular to the line joining the source and a point 25 m from it. Join the vertices of the square to the point source thus forming a pyramid with vertex at the source A. Now consider the section at a distance 5 m from the vertex and parallel to the first square. It will be another square with a side BC. In the similar triangles ABC and ADE,
BC/DE = 5/25 =1/5
→BC = DE/5
So the side of the square at 5 m distance will be one-fifth of the square at 25 m, Hence the area of the square at 5 m will be one-twenty-fifth of the area of the square at 25 m. It is clear from the figure that the same amount of energy (say U) is passing through both areas. If the area of the square at 25 m =A then the area of the square at 5 m =A/25.
The Intensity at 25 m, I = U/A and the intensity at 5 m, I' = U/(A/25) =25(U/A) =25*I
→I =I'/25.
Given that I' = 1.0 x10⁻⁸ W/m²
Hence I = 1.0 x10⁻⁸/25 W/m² = 0.04x10⁻⁸ W/m² =4.0x10⁻¹⁰ W/m²
(It is clear from this diagram that the intensity at a point is inversely proportional to the square of the distance from a point source)
19. The sound level at a point 5.0 m away from a point source is 40 dB. What will be the level at a point 50 m away from the source?
ANSWER: The sound level is given as
ß = 10 log₁₀ [I/I₀]
Given that ß₅ = 40 dB
If the intensity at 5.0 m = I₁ and the intensity at 50 m = I₂, then
The sound level at 5.0 m ß₅ = 10 log₁₀ [I₁/I₀]
The sound level at 50 m ß₅₀ =10 log₁₀ [I₂/I₀]
So, ß₅₀ - ß₅ = 10 log₁₀ [I₂/I₀] - 10 log₁₀ [I₁/I₀]
=[10 log₁₀ I₂ -10 log₁₀ I₀] - [10 log₁₀ I₁ -10 log₁₀ I₀]
= 10 log₁₀ I₂- 10 log₁₀ I₁
= 10 log₁₀ [I₂/I₁]
Since the intensity at a point is inversely proportional to the square of the distance from the point source,
I₂/I₁ = 5²/50² = 1/100
So, ß₅₀ - ß₅ = 10 log₁₀ [I₂/I₁] =10 log₁₀ [1/100]
→ ß₅₀ - ß₅ = 10[log₁₀ 1 -log₁₀ 100] = 10[0 - 2]
→ ß₅₀ - ß₅ = -20
→ ß₅₀ = ß₅ -20 = 40 - 20 = 20 dB
20. If the intensity of sound is doubled, by how many decibels does the sound level increase?
ANSWER: For the intensity I, sound level
ß = 10 log₁₀ [I/I₀]
When the intensity is 2I, sound level
ß' = 10 log₁₀ [2I/I₀]
Hence ß' - ß = 10 log₁₀ [2I/I₀]-10 log₁₀ [I/I₀]
→ß' -ß =10 log₁₀ [2I/I] = 10 log₁₀ 2 =10*0.3 =3 dB
→ß' = ß + 3 dB
So when the intensity is doubled the sound level is increased by 3 dB.
ANSWER: The speed of sound in a gas is given as
V =√{𝜸P/ρ}
Here 𝜸 = Cₚ/Cᵥ =3.5 /2.5 =1.4
P = 1.0 x 10⁵ N/m²
ρ = mass/volume = (32/1000 kg)/(22.4/1000 m³)
=32/22.4 kg/m³
Hence the speed of sound in oxygen
V =√{1.4*1.0x10⁵ *22.4/32} m/s
=√(98000) m/s
=313 m/s
12. The speed of sound as measured by a student in the laboratory on a winter day is 340 m/s when the room temperature is 17 °C. What speed will be measured by another student repeating the experiment on a day when the room temperature is 32 °C?
ANSWER: The speed of sound is proportional to the square root of the temperature, hence
V/V' = √T/√T'
→V' =V√T'/√T
Here speed at 17 °C, V = 340 m/s,
T = 273 + 17 K =290 K
T' = 273 + 32 K =305 K
Speed at 32 °C, V' = 340*√305/√290 m/s
→V' = 340*1.026 m/s =349 m/s
13. At what temperature will the speed of sound be double of its value at 0 °C?
ANSWER: Let the speed of sound be double at t °C than at 0 °C.
T =0 °C =273 K
T' = t °C =273 + t K
speed of sound at 0 °C = V, speed of spond at t °C = V' =2V,
V' = V√T'/√T
→2V = V√(T'/T)
→2 = √(T'/T)
→T'/T =4
→T' =4T =4*273 =1092
→273+t =1092
→t = 1092-273 °C =819 °C
14. The absolute temperature of air in a region linearly increases from T₁ to T₂ in a space of width d. Find the time taken by a sound wave to go through the region in terms of T₁, T₂, d and the speed v of sound at 273 K. Evaluate this time for T₁ = 280 K, T₂ = 310 K, d = 33 m and v = 330 m/s.
ANSWER: The temperature of the air at a distance x from the T₁ end,
Tₓ = T₁ + (T₂ - T₁)x/d
 |
| Diagram for Q-14 |
The speed of sound at x will be
Vₓ = V√Tₓ/√T
where V is the speed of sound at 0 °C.
→Vₓ =V√{T₁+(T₂ - T₁)x/d}/√T
=V√[{T₁d+(T₂ - T₁)x}/Td]
Consider a small distance dx. The small time dt taken by the sound waves to travel the distance dx is given by
dt = dx/Vₓ
→dt = dx/[V√[{T₁d+(T₂ - T₁)x}/Td]
→dt = dx/[V√[{T₁/T+(T₂ - T₁)x/Td]]
Integrating dt for x = 0 to x = d,
t =∫dt =∫dx/[V√[T₁/T+(T₂ - T₁)x/Td]
=2[V√[T₁/T+(T₂ - T₁)x/Td]/[V²(T₂-T₁)/Td]
=2Td√[T₁/T+(T₂ - T₁)x/Td]/V(T₂-T₁)
=2√[Td{T₁d+(T₂ - T₁)x}]/V(T₂-T₁)
=2{√T√d/V}[√{T₁d+(T₂ - T₁)x}/(T₂-T₁)]
putting the integration values
=2{√T√d/V(T₂-T₁)}[√{T₁d+(T₂ - T₁)d} -√(T₁d)]
=2{√T√d/V(T₂-T₁)}[√(T₂d) - √(T₁d)]
=2{√T√d/V(T₂-T₁)}[(√T₂ - √T₁)√d]
=2√T*d(√T₂ - √T₁)/V(T₂ - T₁)
=2√T*d(√T₂ - √T₁)/V{(√T₂)² - (√T₁)²}
=2d√T/V(√T₂ + √T₁)
(Putting value T = 273 K)
=2d√273/V(√T₂ + √T₁)
[NOTE: Here the integration is based on
∫dx/√(a+bx) =2√(a+bx)/b]
For T₁ = 280 K, T₂ = 310 K, d = 33 m and V =330 m/s the time t =2*33*√273/330(√310 +√280) s
=3.30/(17.61+16.73) s
=3.30/34.34 s
=0.096 s
=96 ms
15. Find the change in the volume of 1.0-liter kerosene when it is subjected to an extra pressure of 2.0 x 10⁵ N/m² from the following data. Density of kerosene = 800 kg/m³ and speed of sound in kerosene = 1330 m/s.
ANSWER: The speed of sound V =√(B/ρ)
where B = bulk modulus and ρ = density of the liquid.
From this B =ρV² =800*1330² N/m²
→B = 1.42x10⁹ N/m²
Given volume of Kerosene, v = 1 liter =1x10⁻³ m³
Let the change in volume with the extra pressure =Δv
So, p/(Δv/v) = B, where p = pressure.
→Δv/v = p/B
→Δv = pv/B =2.0x10⁵*1x10⁻³/1.42x10⁹ m³
=1.40x10⁻⁷ m³
=1.40x10⁻⁷*10⁶ cm³
=1.40x0.10 cm³
=0.14 cm³
16. Calculate the bulk modulus of air from the following data about a sound wave of wavelength 35 cm traveling in air. The pressure at a point varies between (1 x 10⁵ 土 14) Pa and the particles of the air vibrate in a simple harmonic motion of amplitude 5.5 x 10⁻⁶ m.
ANSWER: The pressure amplitude p₀ =14 Pa
The displacement amplitude s₀ = 5.5x10⁻⁶ m
Wavelength, λ = 35 cm =0.35 m
Wave number k =2π/λ
The relation between p₀, s₀, k and B is given by
p₀ = Bks₀ =2πBs₀/λ, {Since k = 2π/λ}
→B = p₀λ/2πs₀ =14*0.35/(2π*5.5x10⁻⁶) N/m²
→B = 0.14x10⁶ N/m²
→B = 1.4x10⁵ N/m²
17. A source of sound operates 2.0 kHz, 20 W emitting sound uniformly in all directions. The speed of sound in air is 340 m/s and the density of air is 1.2 kg /m³.
(a) What is the intensity at a distance of 6.0 m from the source?
(b) What will be the pressure amplitude at this point?
(c) What will be the displacement amplitude at this point?
ANSWER: (a) The intensity of sound is the energy passing through a unit area in unit time. Here 20 W i.e. 20 J of energy is passing per second through a spherical surface having radius 6.0 m. Area of this surface =4π*6² m² = 144π m².
Hence the intensity of sound at 6.0 m from the source, I = 20/144π W/m²
= 0.044 W/m²
= 44 mW/m²
(b) The relation between the intensity of sound and the pressure amplitude is given as
I = p₀²/2ρV
→p₀² = 2ρVI
Given that,
V = 340 m/s, ρ = 1.2 kg/m³, from (a) above I =0.044 W/m²
Hence p₀² =2*1.2*340*0.044
→p₀² = 36
→p₀ = 6.0 Pa
(c) Given frequency 𝜈 = 2.0 kHz =2000 Hz
The intensity is also given as
I =p₀²V/2B and the relation between p₀ and s₀ (displacement amplitude) is given as p₀ = B⍵s₀/V i.e. B = p₀V/⍵s₀. Eliminating B, we have
I = p₀²V⍵s₀/2p₀V =p₀⍵s₀/2 =p₀(2π𝜈)s₀/2
→0.044 = 6*(2π*2000)s₀/2
→s₀ = 0.044/12000π m =1.2x10⁻⁶ m
18. The intensity of sound from a point source is 1.0 x10⁻⁸ W/m² at a distance of 5.0 m from the source. What will be the intensity at a distance of 25 m from the source?
 |
| Diagram for Q-18 |
ANSWER: Consider a square of side DE perpendicular to the line joining the source and a point 25 m from it. Join the vertices of the square to the point source thus forming a pyramid with vertex at the source A. Now consider the section at a distance 5 m from the vertex and parallel to the first square. It will be another square with a side BC. In the similar triangles ABC and ADE,
BC/DE = 5/25 =1/5
→BC = DE/5
So the side of the square at 5 m distance will be one-fifth of the square at 25 m, Hence the area of the square at 5 m will be one-twenty-fifth of the area of the square at 25 m. It is clear from the figure that the same amount of energy (say U) is passing through both areas. If the area of the square at 25 m =A then the area of the square at 5 m =A/25.
The Intensity at 25 m, I = U/A and the intensity at 5 m, I' = U/(A/25) =25(U/A) =25*I
→I =I'/25.
Given that I' = 1.0 x10⁻⁸ W/m²
Hence I = 1.0 x10⁻⁸/25 W/m² = 0.04x10⁻⁸ W/m² =4.0x10⁻¹⁰ W/m²
(It is clear from this diagram that the intensity at a point is inversely proportional to the square of the distance from a point source)
19. The sound level at a point 5.0 m away from a point source is 40 dB. What will be the level at a point 50 m away from the source?
ANSWER: The sound level is given as
ß = 10 log₁₀ [I/I₀]
Given that ß₅ = 40 dB
If the intensity at 5.0 m = I₁ and the intensity at 50 m = I₂, then
The sound level at 5.0 m ß₅ = 10 log₁₀ [I₁/I₀]
The sound level at 50 m ß₅₀ =10 log₁₀ [I₂/I₀]
So, ß₅₀ - ß₅ = 10 log₁₀ [I₂/I₀] - 10 log₁₀ [I₁/I₀]
=[10 log₁₀ I₂ -10 log₁₀ I₀] - [10 log₁₀ I₁ -10 log₁₀ I₀]
= 10 log₁₀ I₂- 10 log₁₀ I₁
= 10 log₁₀ [I₂/I₁]
Since the intensity at a point is inversely proportional to the square of the distance from the point source,
I₂/I₁ = 5²/50² = 1/100
So, ß₅₀ - ß₅ = 10 log₁₀ [I₂/I₁] =10 log₁₀ [1/100]
→ ß₅₀ - ß₅ = 10[log₁₀ 1 -log₁₀ 100] = 10[0 - 2]
→ ß₅₀ - ß₅ = -20
→ ß₅₀ = ß₅ -20 = 40 - 20 = 20 dB
20. If the intensity of sound is doubled, by how many decibels does the sound level increase?
ANSWER: For the intensity I, sound level
ß = 10 log₁₀ [I/I₀]
When the intensity is 2I, sound level
ß' = 10 log₁₀ [2I/I₀]
Hence ß' - ß = 10 log₁₀ [2I/I₀]-10 log₁₀ [I/I₀]
→ß' -ß =10 log₁₀ [2I/I] = 10 log₁₀ 2 =10*0.3 =3 dB
→ß' = ß + 3 dB
So when the intensity is doubled the sound level is increased by 3 dB.
===<<<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- 6 - Friction
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For more practice on problems on friction solve these- "New Questions on Friction".
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---------------------------------------------------------------------------------
---------------------------------------------------------------------------------
CHAPTER- 5 - Newton's Laws of Motion
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Click here for → QUESTIONS FOR SHORT ANSWER
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Click here for → Newton's Laws of Motion-Exercises(Q. No. 1 to 12)
Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)
-------------------------------------------------------------------------------
-------------------------------------------------------------------------------
CHAPTER- 4 - The Forces
The Forces-
"Questions for short Answers"
Click here for "The Forces" - OBJECTIVE-I
Click here for "The Forces" - OBJECTIVE-II
Click here for "The Forces" - Exercises
--------------------------------------------------------------------------------------------------------------
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CHAPTER- 3 - Kinematics - Rest and Motion
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Click here for EXERCISES (Question number 1 to 10)
Click here for EXERCISES (Question number 11 to 20)
Click here for EXERCISES (Question number 21 to 30)
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Click here for EXERCISES (Question number 41 to 52)
CHAPTER- 2 - "Vector related Problems"
CHAPTER- 2 - "Vector related Problems"
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