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