Question

One end of a taut string of length 3m along the x axis is fixed at $x = 0$ . The speed of the waves in the string is 100 $m/s$ . The other end is vibrating in the y direction so that stationary waves are set up in the string. The possible waveform(s) of these stationary waves is (are)
(A) $y(t) = A\sin \dfrac{{\pi x}}{6}\cos \dfrac{{50\pi t}}{3}$
(B) $y(t) = A\sin \dfrac{{\pi x}}{3}\cos \dfrac{{100\pi t}}{3}$
(C) $y(t) = A\sin \dfrac{{5\pi x}}{6}\cos \dfrac{{250\pi t}}{3}$
(D) $y(t) = A\sin \dfrac{{7\pi x}}{6}\cos \dfrac{{350\pi t}}{3}$

Answer 1

Hint Node is at the fixed end of the taut string and antinode is at the free end. The angular speed $\omega$ is directly related to the wave number and wave velocity.

Formula used: $l = (2n + 1)\dfrac{\lambda }{4}$ where $n$ are natural numbers i.e. (0,1,2,3…) and $\lambda$ is the wavelength of the wave.
$\Rightarrow k = \dfrac{{2\pi }}{\lambda }$ where $k$ is the wavenumber
$\Rightarrow \omega = vk$ where $\omega$ and $v$ are the angular speed the wave speed (linear) respectively.

Complete step by step answer
We begin with the general equation of a stationary or standing wave. This given as
$\Rightarrow y(t) = A\sin kx\cos \omega t$
The string is said to have a length of $l$ and is fixed at $x = 0$ and the other is vibrating in the vertical direction, i.e. only one end is fixed. Thus $x = 0$ is a node while $x = 3$ is an antinode.
For a stationary wave of this nature, the length is given by
$\Rightarrow l = (2n + 1)\dfrac{\lambda }{4}$ where $n$ are natural numbers i.e. (0,1,2,3…) and $\lambda$ is the wavelength of the wave.
Making $\lambda$ subject of the formula by cross multiplying and dividing both sides by $2n + 1$ , we get
$\Rightarrow \lambda = \dfrac{{4l}}{{2n + 1}}$
Now, $k = \dfrac{{2\pi }}{\lambda }$ .
Substituting the above equation into $k$ and simplifying, we have
$\Rightarrow k = \dfrac{\pi }{{2l}}(2n + 1)$
Finally, we recall the formula for angular speed $\omega$ . This is given as
$\Rightarrow \omega = vk$ where $v$ is the wave speed.
We move ahead to calculate $k$ for different values of $n$ and its corresponding $\omega$ $n = 0$ we have
$\Rightarrow k = \dfrac{\pi }{{2(3)}}$ ……(1)
Since $l = 3$
$\Rightarrow k = \dfrac{\pi }{6}$
The corresponding $\omega$ is
$\Rightarrow \omega = 100\left( {\dfrac{\pi }{6}} \right)$
$\Rightarrow \omega = \dfrac{{50\pi }}{3}$
Similarly for $n = 1$
$\Rightarrow k = \dfrac{\pi }{{2(3)}}[2(1) + 1] = \dfrac{\pi }{{2(3)}}(3)$
$\Rightarrow k = \dfrac{\pi }{2}$
The corresponding $\omega$ is
$\Rightarrow \omega = 100\left( {\dfrac{\pi }{2}} \right)$
$\Rightarrow \omega = 50\pi$
Similarly for $n = 2$
$\Rightarrow k = \dfrac{{5\pi }}{6}$
$\Rightarrow \omega = \dfrac{{250\pi }}{3}$
Similarly for $n = 3$
$\Rightarrow k = \dfrac{{7\pi }}{6}$
$\Rightarrow \omega = \dfrac{{350\pi }}{3}$
We can continue for different values of $n$ , however, we shall skip to $n = 7$ to establish our answer.
For $n = 7$ ,
$\Rightarrow k = \dfrac{{7\pi }}{6}$
$\Rightarrow \omega = \dfrac{{350\pi t}}{3}$
Now, replace the values of the $k$ and $\omega$ in the equation of a standing wave and compare with the options.
For $n = 0$
$\Rightarrow y(t) = A\sin \dfrac{{\pi x}}{6}\cos \dfrac{{50\pi t}}{3}$, which is option A.
Thus, option A is a solution.
For $n = 1$ ,
$\Rightarrow y(t) = A\sin \dfrac{{\pi x}}{2}\cos 50\pi t$ which doesn’t correspond with any of the options. This however, shows that option B is not a solution. Since the value of $k$ and $\omega$ are greater than that of option B.
For $n = 2$ ,
$\Rightarrow y(t) = A\sin \dfrac{{5\pi x}}{6}\cos \dfrac{{250\pi t}}{3}$ which corresponds to option C.
For $n = 3$ ,
$\Rightarrow y(t) = A\sin \dfrac{{7\pi x}}{6}\cos \dfrac{{350\pi t}}{3}$
Finally, for $n = 7$ ,
$\Rightarrow y(t) = A\sin \dfrac{{7\pi x}}{6}\cos \dfrac{{350\pi t}}{3}$
Hence, our answers are A, C, D.

Note
A major point of confusion is identifying the nodes and antinodes. When a stretch string is fixed at only one end and allowed to vibrate at the other as in above, the node is at the fixed end while the antinode is at the free end. However, when the rope is fixed at both ends, the node is at the fixed ends while the antinode lies at the center of the rope, at $x = \dfrac{l}{2}$ . And this configuration uses a different formula for $l$ than the one used above.