Question

A figure shows a snap photograph of a vibrating string at $t=0$. The particle $P$ is observed moving up with velocity $20\sqrt{3}\text{cm/s}$. The tangent at $P$ makes an angle ${60}^{\circ }$ with $x$-axis. Then choose the correct option(s):

• A. Wave is moving along $(+)ve$ $x$-axis.
• B. Wave is moving along $(-)ve$ $x$-axis.
• C. Equation of wave is $y=\left(0.4\text{cm}\right)\sin \left[\right(10\pi {\text{s}}^{-1})t+(\frac{\pi }{2}{\text{cm}}^{-1}x+\frac{\pi }{4}]$.
• D. Equation of wave is $y=\left(0.2\text{cm}\right)\sin \left[\right(10\pi {\text{s}}^{-1})t+(\frac{\pi }{2}{\text{cm}}^{-1})x-\frac{\pi }{2}]$.

Answer 1

Answer: (B), (C)

The correct options are
B Wave is moving along $(-)ve$ $x$-axis.
C Equation of wave is $y=\left(0.4\text{cm}\right)\sin \left[\right(10\pi {\text{s}}^{-1})t+(\frac{\pi }{2}{\text{cm}}^{-1}x+\frac{\pi }{4}]$.

As we know, velocity of particle
$V_p=-V\left(\frac{dy}{dx}\right)$
Here, $V_p$ and slope $(=\frac{dy}{dx})$ both are positive, so $V$ must be negative. Hence the wave is moving along negative $x-$axis.

The general equation of wave moving along negative $x-$ axis is
$y=A\sin (\omega t+Kx+\varphi )$
where all parameters have the usual meaning from the given figure.
$K=\frac{2\pi }{\lambda }=\frac{\pi }{2}{\text{cm}}^{-1}$ $[\because \lambda =4\times {10}^{-2}\text{m}=4\text{cm}$]
$A=4\times {10}^{-3}\text{m}=0.4\text{cm}$
Frequency $\left(f\right)=\frac{V}{\lambda }=\frac{20}{4}=5\text{Hz}$
$\therefore \omega =2\pi f=10\pi \text{rad/s}$

Now from figure at $t=0,x=0;y=2\sqrt{2}\times {10}^{-3}\text{m}$
so $y=A\sin (\omega t-kx+\varphi )$
$\Rightarrow (2\sqrt{2}\times 0.1\text{cm})=\left(0.4\text{cm}\right)\sin \varphi$
$\Rightarrow \sin \varphi =\frac{1}{\sqrt{2}}$
$\Rightarrow \varphi =\frac{\pi }{4}$
Hence, equation of wave is given by
$y=\left(0.4\text{cm}\right)\sin \left[\right(10\pi {\text{s}}^{-1})t+(\frac{\pi }{2}{\text{cm}}^{-1}x+\frac{\pi }{4}]$.
Therefore, options (b) and (c) are correct.