Question

A travelling wave along a stretched string is given by $y=A\sin (kx-\omega t)$. The maximum velocity of a particle is

• A. $A\omega$
• B. $\frac{\omega }{t}$
• C. $\frac{d\omega }{dk}$
• D. $\frac{x}{t}$

Answer 1

Answer: (A)

$A\omega$

Given, the equation of wave is $y=A\sin (kx-\omega t)$.
The velocity of wave is $\frac{dy}{dt}=\frac{d}{dt}\left(A\sin \right(kx-\omega t\left)\right)$
$=-\omega A\cos (kx-\omega t)$
Maximum value of particle velocity is $\omega A$ at $\cos (kx-\omega t)=-1$.