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Exercise Short Questions of Waves, Chapter 8 of 1st Year Physics

 May 01, 2023     1st Year Physics, Exercise Short Questions, Waves     No comments   

Q # 1. What features do longitudinal waves have in common with transverse waves?

Ans. Following features are common in transverse and longitudinal waves:

  • Both are mechanical waves.
  • Particles oscillate about their mean position in both types of waves.
  • Both transport energy from one place to another.
  • Both satisfy the equation: \(v = f\lambda\)

Q # 2. Is it possible for two identical waves travelling in the same direction along a string to give rise to stationary waves?

Ans. No, it is not possible for two identical waves travelling in the same direction along a string to give rise to stationary waves.

For stationary waves, two identical waves must travel in opposite direction.

Q # 3. A wave is produced along a stretched string but some of its particles permanently show zero displacement. What type of wave is it?

Ans. It is a stationary wave and the points are called Nodes.

Q # 4. Explain the terms crest, trough, node and anti-node.

Ans. 

Crest.The portion of the stationary wave above the mean level is called crest.

Trough.The portion of the stationary wave below the mean level is called trough.

Node. The points of zero displacement in stationary waves are called Nodes

Anti-node. The points of maximum displacement in stationary waves are called anti-nodes.

Q # 5. Why should sound travel faster in solids than in gases?

Ans. The formula for speed of sound is:

\[v = \sqrt{\frac{E}{\rho}}\]

where\(E\ \)is Modulus of Elasticity and \(\rho\ \)is Density of medium.

Although the density of solids is greater than the density of gases but the modulus of elasticity for solids is much greater than gases. That's why, sound travel faster in solids than in gases.

Q # 6. How are the beats useful in tuning musical instruments?

Ans. We know that the number of beats produced per second is equal to the difference of frequencies of the two bodies. To tune a musical instrument to the required frequency, it is sounded together with an instrument of known frequency. Now the number of beats produce will tell the difference of their frequency.

The frequency of the untuned instrument is adjusted till the number of beats become zero. At this stage, the two instruments will have the same frequencies. Thus the musical instrument is said to be tuned.

Q # 7. When two notes of frequencies \(\mathbf{f}_{\mathbf{1}}\) \textbf{and} \(\mathbf{f}_{\mathbf{2}}\) are sounded together, beats are formed. If \(\mathbf{f}_{\mathbf{1}}\mathbf{>}\mathbf{f}_{\mathbf{2}}\) what will be the frequency of the beats?
(i) \(\mathbf{f}_{\mathbf{1}}\mathbf{+}\mathbf{f}_{\mathbf{2}}\)
(ii) \(\frac{\mathbf{1}}{\mathbf{2}}\left( \mathbf{f}_{\mathbf{1}}\mathbf{+}\mathbf{f}_{\mathbf{2}} \right)\)
(iii) \(\mathbf{f}_{\mathbf{1}}\mathbf{-}\mathbf{f}_{\mathbf{2}}\)
(iv) \(\frac{\mathbf{1}}{\mathbf{2}}\left( \mathbf{f}_{\mathbf{1}}\mathbf{-}\mathbf{f}_{\mathbf{2}} \right)\)

Ans. As Number of beats per second is equal to the difference between the frequencies of the notes.

 Therefore, the correct answer is (iii) \(f_{1}\mathbf{-}f_{2}\) 

Q # 8. As a result of a distant explosion, an observer senses a ground tremor and then hears the explosion. Explain the time difference.

Ans. The waves produced by the explosion reach the observer quickly through the ground as compared to the sound waves reaching through the air. This is due to the reason that sound travels faster in solid than gases.

Q # 9. Explain why sound travels faster in warm air than in cold air.

Ans. The speed of sound varies directly as the square root of absolute temperature, i.e.,

\[v\ \ \alpha\ \ \ \sqrt{T}\]

It means that greater the temperature of air, more will be the speed of sound in it. That's why sound travel faster in warm air than in cold air.

Q # 10. How should a sound source move with respect to an observer so that the frequency of its sound does not change?

Ans. If the relative velocity between the source and the observer is zero, then there will be no change in frequency of the source and the apparent frequency will be zero.

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