MHT CET · Physics · Waves and Sound
A transverse wave in a medium is given by \(y=A \sin 2(\omega t-k x)\). It is found that the magnitude of the maximum velocity of particles in the medium is equal to that of the wave velocity. What is the value of \(A\) ?
- A \(\frac{2 \lambda}{\pi}\)
- B \(\frac{\lambda}{\pi}\)
- C \(\frac{\lambda}{2 \pi}\)
- D \(\frac{\lambda}{4 \pi}\)
Answer & Solution
Correct Answer
(D) \(\frac{\lambda}{4 \pi}\)
Step-by-step Solution
Detailed explanation
The given equation is \(y=A \sin 2(\omega t-k x)\)
\(\therefore \quad\) Velocity of the particle, \(\mathrm{v}=\frac{\mathrm{dy}}{\mathrm{dt}}\)
\(=2 \mathrm{~A} \omega \cos 2(\omega \mathrm{t}-\mathrm{kx})\)
\(\therefore \quad\) Maximum velocity \(=2 \mathrm{~A} \omega\)
Velocity of the wave \(=\frac{\omega}{\mathrm{k}}\)
Given 2A \(\omega=\frac{\omega}{\mathrm{k}}\)
\(\therefore \quad A=\frac{1}{2 k}=\frac{\lambda}{(2 \pi)^2}=\frac{\lambda}{4 \pi}\)
\(\therefore \quad\) Velocity of the particle, \(\mathrm{v}=\frac{\mathrm{dy}}{\mathrm{dt}}\)
\(=2 \mathrm{~A} \omega \cos 2(\omega \mathrm{t}-\mathrm{kx})\)
\(\therefore \quad\) Maximum velocity \(=2 \mathrm{~A} \omega\)
Velocity of the wave \(=\frac{\omega}{\mathrm{k}}\)
Given 2A \(\omega=\frac{\omega}{\mathrm{k}}\)
\(\therefore \quad A=\frac{1}{2 k}=\frac{\lambda}{(2 \pi)^2}=\frac{\lambda}{4 \pi}\)
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