Question

A transverse sinusoidal wave moves along a string in the positive x-direction at a speed of $10{\text{cm s}}^{-1}$. The wavelength of the wave is $0.5\text{m}$ and its amplitude is $10\text{cm}$. At a particular time $t,$ the snapshot of the wave is shown in the figure. The velocity of the point $P$ when its displacement is $5\text{cm}$ is

• A. $\frac{\sqrt{3}\pi }{50}\hat{i}{\text{ms}}^{-1}$
• B. $-\frac{\sqrt{3}\pi }{50}\hat{i}{\text{ms}}^{-1}$
• C. $-\frac{\sqrt{3}\pi }{50}\hat{j}{\text{ms}}^{-1}$
• D. $\frac{\sqrt{3}\pi }{50}\hat{j}{\text{ms}}^{-1}$

Answer 1

The correct option is D $\frac{\sqrt{3}\pi }{50}\hat{j}{\text{ms}}^{-1}$

General equation of sinusoidal wave is $y=A\sin (\omega t\pm \varphi ).$

Given $A=10\text{cm}$ and $y=5\text{cm}$,

$\therefore \sin (\omega t\pm \varphi )=\frac{1}{2}=\sin {30}^{\circ }$

$\Rightarrow \omega t\pm \varphi ={30}^{\circ }$

Here, $\omega =\frac{2\pi }{T}=2\pi \left(\frac{V}{\lambda }\right)=2\pi \left(\frac{10}{50}\right)=\frac{2\pi }{5}{\text{rad s}}^{-1}$

$\therefore$ The velocity of the point $P$ is
$v_p=\frac{dy}{dt}=A\omega \cos (\omega t\pm \varphi )=\left(10\text{cm}\right)\left(\frac{2\pi }{5}{s}^{-1}\right)\left(\cos {30}^{\circ }\right)$
$v_p=2\pi \sqrt{3}{\text{cm s}}^{-1}=\frac{\pi \sqrt{3}}{50}{\text{ms}}^{-1}$
$\therefore \overrightarrow{v_p}=\frac{\pi \sqrt{3}}{50}\hat{j}{\text{ms}}^{-1}$