Search any question & find its solution
Question:
Answered & Verified by Expert
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 $A$ ?
Options:
Solution:
1193 Upvotes
Verified Answer
The correct answer is:
$\frac{\lambda}{4 \pi}$
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 $\omega=\frac{\omega}{\mathrm{k}}$
$\therefore \quad A=\frac{1}{2 k}=\frac{\lambda}{(2 \pi)^2}=\frac{\lambda}{4 \pi}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.