MHT CET · Physics · Oscillations
A particle performs S.H.M. of amplitude 'A' and wavelength ' \(\lambda\) ', Then the velocity of the wave \((\mathrm{V})\) and the maximum particle velocity \((\nu)\) are related as
- A \(\nu=\frac{\lambda v}{4 \pi A}\)
- B \(\mathrm{V}=\frac{\lambda \nu}{4 \pi \mathrm{~A}}\)
- C \(v=\frac{2 \pi A}{\lambda} V\)
- D \(V=\frac{2 \pi A}{\lambda} v\)
Answer & Solution
Correct Answer
(C) \(v=\frac{2 \pi A}{\lambda} V\)
Step-by-step Solution
Detailed explanation
\(v = A\omega\) \(\omega = 2\pi f\)
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