MHT CET · Physics · Oscillations
A simple harmonic progressive wave is represented as \(Y=A \sin 2 \pi\left(n t-\frac{x}{\lambda}\right) \mathrm{cm}\). If the maximum particle velocity is four times the wave velocity, then the wavelength of the wave is
- A \(\frac{\pi \mathrm{A}}{4}\)
- B \(4 \pi \mathrm{A}\)
- C \(2 \pi \mathrm{A}\)
- D \(\frac{\pi \mathrm{A}}{2}\)
Answer & Solution
Correct Answer
(D) \(\frac{\pi \mathrm{A}}{2}\)
Step-by-step Solution
Detailed explanation
\(v_{p,max} = A\omega = A(2\pi n)\) \(v_{wave} = n\lambda\)
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