Question

A transverse sinusoidal wave moves along a string in the positive $x-$direction at a speed of $10\text{cm/s}$. The wavelength of the wave is $0.5m$ and its amplitude is $10cm$. At a particular time $t$, the snap-shot of the wave is shown in figure. The velocity of point $P$ when its displacement is $0.05m$ is

• A. $\frac{\sqrt{3}\pi }{50}\hat{j}\text{m/s}$
• B. $-\frac{\sqrt{3}\pi }{50}\hat{j}\text{m/s}$
• C. $\frac{\sqrt{3}\pi }{50}\hat{i}\text{m/s}$
• D. $-\frac{\sqrt{3}\pi }{50}\hat{i}\text{m/s}$

Answer 1

The correct option is B $-\frac{\sqrt{3}\pi }{50}\hat{j}\text{m/s}$


Since the wave is progressing along the positive x-direction, the velocity of the particle at $P$ has to be downward ($-\hat{j}$ direction)

Each particle executes SHM about its mean positions. So we can find velocity using the relation.

$v=\omega \sqrt{A^2-y^2}$
$=\omega \sqrt{{0.1}^2-{0.05}^2}$
$=\omega \sqrt{{0.1}^2-{0.05}^2}$

($\omega =k\ast v_{wave}=\frac{2\pi }{\lambda }\ast v_{wave}=\frac{2\pi }{0.5}\ast 0.1=0.4\pi s^{-1}$

$\Rightarrow v=0.4\pi \ast 0.05\ast \sqrt{2^2-1}=\sqrt{3}\pi \ast 0.02=\frac{\sqrt{3}\pi }{50}$

Therefore, velocity of the particle at P is $-\frac{\sqrt{3}\pi }{50}\hat{j}$