Question

One end of a taut string of length $3\text{m}$ along the $\text{x-axis}$ is fixed at $x=0$. The speed of the waves in the string is $100\text{m/s}$. The other end of the string is vibrating in the $\text{y-direction}$ so that stationary waves are set up in the string. The possible waveform $($s$)$ of these stationary wave is $($are$)$

• A. $y\left(t\right)=A\sin \frac{\pi x}{6}\cos \frac{50\pi t}{6}$
• B. $y\left(t\right)=A\sin \frac{\pi x}{3}\cos \frac{100\pi t}{3}$
• C. $y\left(t\right)=A\sin \frac{5\pi x}{6}\cos \frac{250\pi t}{3}$
• D. $y\left(t\right)=A\sin \frac{5\pi x}{2}\cos 250\pi t$

Answer 1

Answer: (C), (D)

$y\left(t\right)=A\sin \frac{5\pi x}{2}\cos 250\pi t$

We know that fixed end forms a node while the free end forms an antinode.

Therefore, at $x=0$ is a node and at $x=3\text{m}$ is an antinode.

Possible modes of vibration of a string fixed at one end are

$L=(2n+1)\frac{\lambda }{4}$ where, $n=0,1,2,\mathrm{3...}$

or, $\lambda =\frac{4L}{2n+1}$

or , $\lambda =\frac{12}{2n+1}(\because L=3\text{m})$

We know that,
$k=\frac{2\pi }{\lambda }$

$\Rightarrow k=\frac{2\pi }{12/(2n+1)}$

or , $k=\frac{(2n+1)\pi }{6}$

Now, we know that , $\omega =vk$

$\therefore \omega =100(2n+1)\frac{\pi }{6}$

or , $\omega =\frac{50\pi (2n+1)}{3}$

For, $n=0$, $k=\frac{\pi }{6}$ and $\omega =\frac{50\pi }{3}$

For, $n=1$, $k=\frac{\pi }{2}$ and $\omega =50\pi$

For, $n=2$, $k=\frac{5\pi }{6}$ and $\omega =\frac{250\pi }{3}$

Similarly,
For, $n=7$, $k=\frac{5\pi }{2}$ and $\omega =250\pi$

and so on.

Now, equation of a standing wave for a string fixed at one end is given by

$y(x,t)=A\sin \left(kx\right)\cos \left(\omega t\right)$

For $n=0,$ $y(x,t)=A\sin \frac{\pi x}{6}\cos \frac{50\pi t}{3}$

For $n=1,$$y(x,t)=A\sin \frac{\pi x}{2}\cos 50\pi t$

For $n=2,$$y(x,t)=A\sin \frac{5\pi x}{6}\cos \frac{250\pi t}{3}$

Similarly,

For $n=7,$ $y(x,t)=A\sin \frac{5\pi x}{2}\cos 250\pi t$

From this we can conclude that, options (c) and (d) are the correct answers.