Chapter 5 Sound Long Answer Type Question And Answers
Question 1. Differentiate between transverse and longitudinal waves.
Answer.
Difference between transverse and longitudinal waves:
Question 2. Explain how defects in a metal block can be detected using ultrasound.
Answer.
Defects in a metal block can be detected using ultrasound:
Ultrasounds are used to detect cracks in the metal blocks. The cracks inside the metal blocks, which are imperceptible from outside, decreases the strength of the structure.
Ultrasonic waves can pass through the metal block and detectors are conditioned to detect the transmitted waves. If there is even a small defect, the ultrasound gets reflected indicating the occurrence of the flaw or defect in that metal block.
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Question 3. Calculate the difference in Newton’s value of velocity of sound in air and that found by Pascal.
Answer.
The difference in Newton’s value of velocity of sound in air and that found by Pascal:
Velocity of sound in a gas is given by
v = \(\sqrt{\frac{E}{\rho}}\)
Where E is the modulus of elasticity of the gas.
Newton said ‘when sound travels through air through compression and rarefactions, the changes taking place in air are isothermal changes.’ For isothermal changes,
E = Pressure of gas
= Pressure of air
= 1.013 × 105 Nm-2
∴ According to Newton,
va = \(\sqrt{\frac{P_a}{\rho_a}}\)
va = \(\sqrt{\frac{1.013 \times 10^5 \mathrm{Nm}^{-2}}{1.29 \mathrm{kgm}^{-3}}}\)
Newton’s value of va = 280 ms-1
Almost a century later, Laplace said ‘Sound travels very rapidly through air such rapid changes cannot be isothermal. They must be adiabatic in nature.’
For adiabatic changes,
E = YP
= 1.4 × 1.013 × 105 Pa
So, according to Pascal,
va = \(\sqrt{\frac{Y P_a}{\rho_e}}\)
= \(\sqrt{\frac{1.4 \times 1.013 \times 10^5}{1.29}}\)
= 331.6 ms-1
Velocity of sound in air at 0 °C is about 332 ms-1.
So, Laplace’s value is very close to the actual value of velocity of sound in air. Thus Laplace corrected Newton. This is therefore, called
Laplace’s correction.
∆v = (331.6 − 280)ms-1.
= 51.6 ms-1.
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Question 4. At what temperature the velocity of sound in air, will be double that of air at 0 °C ? (Given v0 = 332 ms-1).
Answer.
We know that,
\(\frac{v_2}{v_1}=\sqrt{\frac{T_2}{T_1}}\) \(\frac{v_2}{v_0}=\sqrt{\frac{T_2}{T_1}}\)\(\frac{2 v_0}{v_0}=\sqrt{\frac{T_2}{273}}\) (∴ V2 = 2V0)
Squarring on both sides, we get
\(4=\frac{T_2}{273}\)T2 = 1092 K
= (1092 − 273) °C
= 819 °C
At 819 °C temperature the velocity of sound in air, will be double that of air at 0 °C
Question 5. Velocity of sound in a gas at STP is 400 ms-1. If both the pressure and the absolute temperature of the gas are doubled, then how will the velocity of sound change?
Answer.
Given:
Velocity of sound in a gas at STP is 400 ms-1. If both the pressure and the absolute temperature of the gas are doubled
Change in pressure has no effect on the velocity of sound. Only temperature change has,
Given: T1 = 0 °C = 273 K, v1 = 400 ms-1, T2 = 2 × 273 k,
We know that, T2 = ?
\(\frac{v_2}{v_1}=\sqrt{\frac{T_2}{T_1}}\)\frac{v_2}{400}=\sqrt{\frac{(2 \times 273)}{273}}
v2 = 400 \sqrt{2}
= 565.7 ms-1.
The Change In Velocity Of Sound Is 565.7 ms-1.
Question 6. A boy blows a whistle and its echo from a distant wall is heard after 1.8 seconds. How far is the wall from him? Velocity of sound in air at room temperature = 340 ms-1.
Answer.
Given:
A boy blows a whistle and its echo from a distant wall is heard after 1.8 seconds.
If d be the distance of the wall from the boy, then sound travels a distance d to the wall and back to the boy.
So, total distance travelled by sound is 2d.
Time taken by sound to go to the wall(reflection of sound) and back.
\(t=\frac{\text { Distance }}{\text { Velocity }}\) \(t=\frac{2 d}{v}\)=> \(1.6=\frac{2 d}{340}\)
2d = 340 x 1.6
∴ \(d=\frac{340 \times 1.6}{2}\)
= 272 m
Question 7. Two ships are 6 km apart in Indian Ocean. Signal sent by one ship is reflected by the ocean bed and received at the second ship after 6.6 s. Find the depth of the Indian Ocean. (Velocity of sound in water = 1500 ms-1)
Answer.
Given:
Two ships are 6 km apart in Indian Ocean. Signal sent by one ship is reflected by the ocean bed and received at the second ship after 6.6 s.
The figure below shows two ships S1 and S2. Signal sent by ships S1 travels to the ocean bed along S1B and gets reflected along BS2 to reach the ship S2. So, distance travelled by sound is S1BS2 = 2 S1B.
In ∆ S1BM,
\(S_1 B=\sqrt{d^2+\left(S_1 M\right)^2}\)Time taken by (sound) signal to travel distance x = S1BS2 is given by
\(t=\frac{\text { Distance }}{\text { Velocity }}=\frac{S_1 B S_2}{v}\)= \(\frac{S_1 B+B S_2}{v}\)
= \(\frac{2 S_1 B}{v}\) (∴ Bs2 = s1B)
\(t=\frac{2 \sqrt{d^2+\left(S_1 M\right)^2}}{v}\)∴ \(\sqrt{d^2+\left(S_1 M\right)^2}=\left(\frac{v t}{2}\right)\)
Squarring on both sides,
\(d^2+(S, M)^2=\left(\frac{v t}{2}\right)^2\) \(d^2+(3000)^2=\left(\frac{1500 \times 6.6}{2}\right)^2\)d2 + (3000)2 = (4950)2
∴ d2 = 49502 − 30002
= (4950 + 3000)(4950 − 3000)
d2 = (7950)(1950)
\(d=\sqrt{(7950) \times(1950)}\)⇒ d = 3937.4 m
⇒ d ≅ 3940 m
⇒ d ≅ 3.94 km
The depth of the Indian Ocean ≅ 3.94 km
Question 8. A sound wave of wavelength 0.332 m has a time period of 10-3 s. If the time period is decreased to 10-4 s. Calculate the wavelength and frequency of the new wave. Name the subjective property of sound related to its frequency and the light related to its wavelength.
Answer.
Given:
A sound wave of wavelength 0.332 m has a time period of 10-3 s. If the time period is decreased to 10-4 s.
Here λ = 0.332 m
Time taken to travel,
t = 10-3 s
Velocity = \(\frac{\lambda}{t}\)
v = \(\frac{0.33}{10^{-3}}\)
v = 0.33 × 103
v = 330 m/s
Time period for second wave = 10-4 s
Wavelength λ = vT
= 330 × 10-4
Wavelength λ = 0.033 m
Frequency = \(\frac{1}{T}\)
= \(\frac{1}{10^{-3}}\)
Frequency = 103 Hz
Question 9. Derive a relation between Wave-Velocity, frequency and wavelength.
Answer.
A relation between Wave-Velocity, frequency and wavelength:
Let velocity of wave = v
Time period = T
Frequency = υ
Wavelength = λ
As per the definition of wavelength,
λ = Distance travelled by wave in one time period
= Wave Velocity × Time period
= v × T
vT = λ
t = \(\frac{1}{v}\)
\(v \times\left(\frac{1}{v}\right)=\lambda\)Therefore, v = λυ.
Hence,
Wave Velocity = Frequency × Wavelength
Question 10. Meera is standing between two hills. She shouted loudly and hears first echo after 0.5 sec and second echo after 1 sec. What is the distance between two hills?
Answer.
Given:
Meera is standing between two hills. She shouted loudly and hears first echo after 0.5 sec and second echo after 1 sec.
Let the distance between nearest cliff and
Meera = x m and the distance between distant cliff and Meera = y m
Distance between two cliff = (x + y) m
Total distance covered by sound to produce first echo = 2x m and time = 0.5 sec.
2x = 340 × 0.5
\(x=340 \times \frac{0.5}{2}=85 \mathrm{~m}\)Total distance covered by sound to produce second echo = 2y m and time = 1 sec
2y = 340 × 1
y = \(340 \times \frac{1}{2}\)
y = 170m
So distance between the two cliffs = (85 + 170) = 255 m
Question 11. A man fires a gun and hears its echo after 5 seconds. The man then moves 310 m towards the hill and fires his gun again. This time he hears the echo after 3 seconds. Calculate the speed of sound.
Answer.
Given:
A man fires a gun and hears its echo after 5 seconds. The man then moves 310 m towards the hill and fires his gun again. This time he hears the echo after 3 seconds.
Let d be the distance between the man and the hill in the beginning
v = \(\frac{2 d}{t}\)
v = \(\frac{2 d}{5}\) (1)
He moves 310 m towards the hill. Therefore distance will be (d − 310) m.
v = \(2 \frac{(d-310)}{3}\) (2)
Since velocity of sound is same, equating (1) and (2).
\(\frac{2 d}{5}=2 \frac{(d-310)}{3}\)3d = 5d − 1550
2d = 1550
d = 775 m
Velocity of sound v = \(2 \times \frac{775}{5}\)
Velocity of sound v = 310 ms-1
Question 12. An engine is approaching a hill at a constant speed. When it is at a distance 0.9 km, it blows whistles, whose echo is heard by the driver after 5 s. If the speed of sound is 340 ms-1,calculate the speed of the engine.
Answer.
Given:
An engine is approaching a hill at a constant speed. When it is at a distance 0.9 km, it blows whistles, whose echo is heard by the driver after 5 s.
Let ve be the speed of the engine.
Distance covered by the engine in 5 s = 5 ve
Distance covered by sound in reaching the hill and coming back to moving driver
= 900 × 2 − 5ve
= 1800 − 5ve
According to given condition,
t = \(\frac{\left(1800-5 v_e\right)}{v}\)
As t = 5 s and v (speed of sound) = 340 ms-1
5 = \(\frac{(1800-5 v)}{340}\)
1700 = 1800 − 5ve
Or ve = 20 ms-1