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)

$\frac {\lambda}{4 \pi}$

Answer 2

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)
$\therefore \frac{v}{2} = \frac{2\pi va}{\lambda}$
$\Rightarrow a = \frac{\lambda}{4\pi}$