Question

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

• A. $A\omega$
• B. $\omega /k$
• C. $d\omega /dk$
• D. $x/l$

Answer 1

The correct option is A $A\omega$

Given that, the displacement of a particle is
$y=A\sin (\omega t-kx)$ ......(i)
The particle velocity
$v_p=\frac{dy}{dt}$ ......(ii)
Now, on differentiating equation (i) with respect to, $t$,
$\frac{dy}{dt}=A\cos (\omega t-kx)\cdot \omega$
$\Rightarrow \frac{dy}{dt}=A\omega \cos (\omega t-kx)$
From equation (ii)
$\Rightarrow v_p=A\omega \cos (\omega t-kx)$
For maximum particle velocity,
$\cos (\omega t-kx)=1$
So, $v_p=A\omega \times 1$
$\Rightarrow v_p=A\omega$