Question

In a transverse progressive wave of amplitude $A$ , the maximum particle velocity is four times its 'wave velocity' then the wavelength of the wave is

• A. $2 \pi A$
• B. $\pi A$
• C. $\frac{\pi A}{2}$
• D. $\frac{\pi A}{4}$

Answer 1

Answer: (C)

Given, Amplitude of the wave $=A$
Maximum velocity of particle $V _{\max }=4 V$
(where $v$ is the velocity of wave)
Maximum velocity of wave, $v_{\max }=4 v$
or $v _{\max }= A \omega=4 v$
$\therefore \omega=\frac{4 V }{ A }$
Wavelength of the wave,
$\lambda=\frac{\text { velocity of wave }( v )}{\text { frequency of wave }( f )}=\frac{ v }{\omega}=\frac{2 \pi v }{\omega} $
$=\frac{2 \pi v }{4 v / A }=\frac{\pi A }{2} \ldots . .\left(\text { where, } f =\frac{\omega}{2 \pi}\right)$