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 v a}{\lambda} $
but $v_{max } = \frac{v}{2} $ (given)
$\therefore \frac{v}{2} = \frac{2\pi va}{\lambda}$
$ \Rightarrow a = \frac{\lambda}{4\pi} $