Question:
In a transverse wave the distance between a crest and neighbouring trough at the same instant is 4.0 cm and the distance between a crest and trough at the same place is 1.0 cm. The next crest appears at the same place after a time interval of 0.4 s. The maximum speed of the vibrating particles in the medium is:
In a transverse wave the distance between a crest and neighbouring trough at the same instant is 4.0 cm and the distance between a crest and trough at the same place is 1.0 cm. The next crest appears at the same place after a time interval of 0.4 s. The maximum speed of the vibrating particles in the medium is:
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Crest-to-trough distance at the same place equals \(2A\); the period gives \(\omega\). Then \(v_{max}=A\omega\).
Updated On: Nov 4, 2026
- \(\frac{3\pi}{2}\) cm/s
- \(\frac{5\pi}{2}\) cm/s
- \(\frac{\pi}{2}\) cm/s
- \(2\pi\) cm/s
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The Correct Option is B
Solution and Explanation
The crest-to-trough separation at the same place is \(2A = 1.0\ \mathrm{cm}\), so \(A = 0.5\ \mathrm{cm}\).
The next crest appears after \(T = 0.4\ \mathrm{s}\), so \(\omega = \dfrac{2\pi}{T} = 5\pi\ \mathrm{rad/s}\).
(The 4.0 cm crest-trough distance only gives \(\lambda = 8\) cm; it is not needed for the particle speed.)
Maximum particle speed \(v_{max} = A\omega = 0.5\times 5\pi = \dfrac{5\pi}{2}\ \mathrm{cm/s}\).
Answer: (2) \(\dfrac{5\pi}{2}\) cm/s
The next crest appears after \(T = 0.4\ \mathrm{s}\), so \(\omega = \dfrac{2\pi}{T} = 5\pi\ \mathrm{rad/s}\).
(The 4.0 cm crest-trough distance only gives \(\lambda = 8\) cm; it is not needed for the particle speed.)
Maximum particle speed \(v_{max} = A\omega = 0.5\times 5\pi = \dfrac{5\pi}{2}\ \mathrm{cm/s}\).
Answer: (2) \(\dfrac{5\pi}{2}\) cm/s
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