Sound Waves Questions

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Very Short Answers Questions [1 Mark Questions]

Ques. What is meant by a sound?

Ans. Sound waves are produced by a vibrating body is a combination of low and high-pressure waves. The vibration causes the particle in the propagation medium to vibrate, causing energy to be transmitted through the surrounding media.

Ques. Define the frequency of a sound wave.

Ans. The frequency of a sound wave is defined as the number of compressions and rarefactions that occur per unit of time.

Ques. What is the impact of temperature on the sound’s speed?

Ans. The speed of sound increases as the temperature of the medium rises. When the temperature of the air rises by 10°C, the speed of the air increases by 0.61 m/s.

Ques. Why are sound waves called longitudinal waves?

Ans. The vibration of the particles moves parallel to the direction of wave propagation. As a result, sound waves are known as longitudinal waves.

Ques. Define the speed of sound.

Ans. The speed of a sound wave is the distance traveled by this acoustic wave per unit of time as it passes through a medium.

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Short Answers Questions [2 Marks Questions]

Ques. Why do sound waves travel fastest in solids?

Ans. Particles in solids are more densely packed than in gases and liquids. As a result, particles within them can collide easily. Because sound is simply a series of particle collisions, it may travel the quickest through solids.

Ques. Sound waves are called mechanical waves. Why?

Ans. Mechanical waves are those that require a medium to transport their energy from one location to another. Sound cannot propagate in a vacuum because it requires a medium. Therefore, sound waves are called mechanical waves.

Ques. What is the speed of sound in the air?

Ans. The speed of sound in air is approximately 343 m/s at 20°C. It is directly dependent on the temperature of the medium as well as the type of media through which the acoustic wave is traveling. At 0°C, its speed is approximately 331 m/s.

Ques. Which characteristic of sound differentiates a shrill or sharp sound from a dull or grave sound?

Ans. Pitch is the quality of a sound that determines a shrill or sharp sound from a dull or grave one. It is entirely dependent on frequency. The greater the frequency of the wave, the higher the pitch and the shriller the sound.

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Long Answers Questions [3 Marks Questions]

Ques. What are the main factors that control the characteristics of the propagation of sound?

Ans. The pressure and density of the given medium have a reciprocal relationship. 

• This relationship is affected by temperature and determines the speed of sound within the medium. 
• Another important aspect of the propagation of sound is the mobility of the medium. 
• When the medium is moving, the average speed of the sound may drop or increase depending on the direction of the movement. 
• The viscosity of the medium is also a significant factor. 
• The viscosity determines the rate at which the intensity of sound waves decreases as they travel away from the source.

Ques. Give an example of how the speed of the medium affects the propagation of sound.

Ans. The speed of the medium affects the propagation of sound because

• When sound waves and wind move in the same direction, the propagation speed of the sound waves is increased with the speed of the wind. 
• On the other hand, if the wind and sound waves are traveling in opposing directions, the speed of sound will decrease as the speed of the wind increases.

Ques. If the speed of the sound wave in water is 1400 m/s, then calculate the bulk modulus of the elasticity of the water. (Density of water is 1 g/cm³)

Ans. Given

• Speed of the sound wave, v = 1400 m/s
• The density of water, ρ = 1 g/cm³ = 1000 kg/m³

The formula for the speed of sound in a medium is given by

v = √(B/ρ)

⇒ B = ρv²

Where ρ is the density of the medium

On substituting the values, we get

B = 1000 x 1400²

⇒ B = 1.96 x 10⁹ N/m²

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Very Long Answers Questions [5 Marks Questions]

Ques. A fighter jet (subsonic) flies overhead at an altitude of 100 m. The intensity of sound is observed to be 150 dB. At what altitude should this plane fly so that the intensity drops to the level of the threshold of pain (i.e. 120 dB)? Ignore the finite time required for the sound to reach the ground.

Ans. We know for every 10 dB increase in loudness, the intensity will increase by 10 times. Therefore, for every 10 dB decrease in loudness, the intensity will decrease by (1/10)^th times.

Here loudness drops to (150 - 120) dB = 30 dB = (3 x 10) dB

Therefore intensity will drop to (1/10 x 1/10 x 1/10 = 1/1000 times) the initial value.

Now let the intensity be I, then the final intensity will be (I/1000).

We know, the intensity of sound is inversely proportional to the square of the distance from the source. i.e. I ∝ 1/r²

∴ I₁ / I₂ = (1/r₁²) / (1/r₂²)

Given

• The initial intensity of the sound, I₁ = I
• The final intensity of the sound, I₂ = I/1000
• Initial altitude, r₁ = 100 m

∴ I / (I/1000) = (r₂/100)²

⇒ 1000 = (r₂²/104)

⇒ r₂² = 107

⇒ r² = 3162 m

Therefore, the altitude should this plane fly so that the intensity drops to the level of the threshold of pain (i.e. 120 dB) is 3162 m.

Ques. The loudness of a source of sound at a given location increases by 10 dB. By how many times does its intensity increase?

Ans. Let L be the initial loudness of the sound wave. Then the initial intensity (I) of the sound is given by

L = 10 log₁₀ (I/I₀)

Where I0 is the threshold of hearing.

Let, when loudness is increased by 10 dB, the intensity becomes nI. Therefore

L + 10 = 10 log₁₀ (nI/I₀)

⇒ L + 10 = 10 [log₁₀ n + log₁₀ (I/I₀)]

⇒ L + 10 = 10 log₁₀ n + 10 log₁₀ (I/I₀)

But 10 log₁₀ (I/I₀) = L, therefore

L + 10 = 10 log₁₀ n + L

⇒ log₁₀ n = 10/10

⇒ log₁₀ n = 1

⇒ n = 10

Therefore for every 10 dB increase in loudness, the intensity will be increased by 10 times.

Ques. The pressure variation in a sound wave is given by

ΔP = 8 cos (4.00x - 3000t + π/4)

Find its displacement amplitude. The density of the medium is 10³ kg/m³.

Ans. The given pressure variation equation is

ΔP = 8 cos (4.00x - 3000t + π/4)

Comparing the above equation with the pressure variation equation

ΔP = ΔP₀ cos (kx - ωt + Φ)

We get

• Pressure amplitude, ΔP₀ = 8
• Propagation constant, k = 4
• Angular frequency, ω = 3000

Pressure amplitude is also given by ΔP₀ = Bak

⇒ a = ΔP₀/Bk …(i)

Where

• B is the bulk modulus
• a is displacement amplitude
• k is propagation constant

The velocity of sound in a medium is given by

v = √(B/ρ)

⇒ B = ρv²

Where ρ is the density of the medium

Also, v = ω/k

⇒ B = ρ(ω/k)²

Substituting the above expression in equation (i), we get

Displacement amplitude, a = ΔP₀ / ρ(ω/k)²k = ΔP₀k/ρω²

On substituting the values, we get

a = (8 x 4) / (1000 x 3000²)

⇒ a = 3.55 x 10^-9 m

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