Question

A transverse wave is passing through a medium. The maximum speed of the vibrating particle occurs when the displacement of the particle from the mean position is:

• A. zero
• B. half of the amplitude
• C. equal to the amplitude
• D. none of these

Answer 1

The correct option is A zero

Let the transverse wave can be given as,
$y=A\sin (\omega t-kx+\varphi )$
We know velocity of particle,
$v_P=\frac{\partial y}{\partial t}=A\omega \cos (\omega t-kx+\varphi )$
For maximum particle velocity,
When $\cos (\omega t-kx+\varphi )=1$
$(v_P{)}_{\text{max}}=A\omega$
$\Rightarrow \sin \theta$ is zero when $\cos \theta =1$
i.e $\sin (\omega t-kx+\varphi )=0$
Substituting it in wave equation we get,
$\Rightarrow y=0$
It represents particle's displacement from its mean position is zero.

ALTERNATIVE:
since the particles oscillate simple harmonically in a transverse wave. The velocity is maximum at the mean position, where net force is zero.
$\Rightarrow$ For maximising $v$,
$\frac{dv}{dt}=0$
i.e $a=0$
$\Rightarrow F_{\text{net}}=0$
Hence particle is at its equilibrium position(Mean position)
Hence displacement from mean position is zero.