Question

The maximum particle velocity in a wavemotion is half the wave velocity. Then the amplitude of the wave is equal to

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

Answer 1

Answer: (D)

For a wave,
$y=a\sin \frac{2\pi }{\lambda }\left(vt-x\right)\dots (i)$
Differentiating Eq (i) w.r.t. t, we get
$\frac{dy}{dt}=\frac{2\pi va}{\lambda }\cos \frac{2\pi }{\lambda }\left(vt-x\right)$
Now, maximum velocity is obtained when
$\cos \frac{2\pi }{\lambda }\left(vt-x\right)=1$
$\therefore v_{max}={\left(\frac{dy}{dt}\right)}_{max}=\frac{2\pi va}{\lambda }$
but $v_{max}=\frac{v}{2}$ (given)
$\therefore \frac{v}{2}=\frac{2\pi va}{\lambda }$
$\Rightarrow a=\frac{\lambda }{4\pi }$