Question

The sketch in the figure shows the displacement time curve of a sinusoidal wave at $x=8m$ . Taking the velocity of the wave $v=6m{s}^{-1}$ along the positive X-axis, write the equation of the wave.
(A) $y=0.5\sin \left({\displaystyle \frac{\pi }{3}}t+{\displaystyle \frac{\pi }{18}}x+{\displaystyle \frac{7\pi }{9}}\right)$
(B) $y=0.5\sin \left({\displaystyle \frac{\pi }{3}}t+{\displaystyle \frac{\pi }{18}}x+{\displaystyle \frac{11\pi }{9}}\right)$
(C) $y=0.5\sin \left({\displaystyle \frac{\pi }{3}}t-{\displaystyle \frac{\pi }{18}}x+{\displaystyle \frac{11\pi }{9}}\right)$
(D) $y=0.5\sin \left({\displaystyle \frac{\pi }{3}}t-{\displaystyle \frac{\pi }{18}}x+{\displaystyle \frac{7\pi }{9}}\right)$

Answer 1

Hint :Use the general equation of a sinusoidal wave and find the different parameters of the wave. The general equation of a sinusoidal wave propagating along positive X-direction is given by, $y=A\sin \left(\omega t-kx+\phi \right)$ . Where, $A$ is the amplitude of the wave, $\omega$ is the angular frequency of the wave, $k$ is the propagation constant of the wave, $\phi$ is the initial phase, $x$ is the position of the wave and $t$ is the time.

Complete Step By Step Answer:
We know that the equation of a sinusoidal wave propagating along positive X- direction is given by, $y=A\sin \left(\omega t-kx+\phi \right)$ . Where, $A$ is the amplitude of the wave, $\omega$ is the angular frequency of the wave, $k$ is the propagation constant of the wave, $\phi$ is the initial phase. $x$ is the position of the wave and $t$ is the time.
From the figure we can see that at $t=2s$ to $t=8s$ the wave completes one single oscillation. Now we know, the time period of a wave is the time required to complete one complete oscillation is called the time period of the wave. Hence the time period of the wave will be, $T=(8-2)=6s$ .
Now, we know that the frequency of a wave is the total number of complete oscillation in one second.
Hence, frequency of the wave will be, $f={\displaystyle \frac{1}{T}}={\displaystyle \frac{1}{6}}Hz$ .
So, the angular frequency of the wave will be, $\omega =2\pi f$ .
Putting the values we get, $\omega =2\pi {\displaystyle \frac{1}{6}}={\displaystyle \frac{\pi }{3}}rad/s$
Now, From the figure we can see that the amplitude of the wave is , $A=0.5cm=5\times {10}^{-3}m$ .
Now we know that the propagation constant of the wave is related to the wavelength of the wave as, $k={\displaystyle \frac{2\pi }{\lambda }}$ . And we know, wavelength of wave is the distance covered in one complete oscillation , $\lambda ={\displaystyle \frac{v}{f}}$ where $v$ is the velocity of the wave.
We have given here that the velocity of the wave is, $v=6m{s}^{-1}$ and we have found $f={\displaystyle \frac{1}{6}}Hz$ .
Putting the values we get, $\lambda ={\displaystyle \frac{6}{{\displaystyle \frac{1}{6}}}}=36m$ .
Hence propagation constant will be, $k={\displaystyle \frac{2\pi }{\lambda }}={\displaystyle \frac{2\pi }{36}}={\displaystyle \frac{\pi }{18}}{m}^{-1}$
So, putting these values in the sinusoidal wave equation we get the wave equation as,
 $y=0.5\sin \left({\displaystyle \frac{\pi }{3}}t-{\displaystyle \frac{\pi }{18}}x+\phi \right)$ [Taking displacement in $cm$ ]
Since, there is insufficient data to find the initial phase of the wave we have to check the available options.
Now, putting, $\phi ={\displaystyle \frac{11\pi }{9}}$ at $t=0$ and $x=8$
we get, $y=0.5\sin (0-{\displaystyle \frac{8\pi }{18}}+{\displaystyle \frac{11\pi }{9}})$
Or, $y=0.5\sin ({\displaystyle \frac{7\pi }{9}})=0.321$
So, putting $\phi ={\displaystyle \frac{7\pi }{9}}$ at $t=0$ and $x=8$
We get, $y=0.5\sin (0-{\displaystyle \frac{8\pi }{18}}+{\displaystyle \frac{7\pi }{9}})$
Or, $y=0.5\sin ({\displaystyle \frac{\pi }{3}})=0.433$
From observation of the graph we can see that the value of $y$ t $t=0$ and $x=8$ must be near the $y⩾0.4cm$ range. Hence, $\phi ={\displaystyle \frac{7\pi }{9}}$ must be the initial phase.
So, equation of the wave will be, $y=0.5\sin \left({\displaystyle \frac{\pi }{3}}t-{\displaystyle \frac{\pi }{18}}x+{\displaystyle \frac{7\pi }{9}}\right)$
So, Option (D) is correct.

Note :
 $\bullet$ Here, if the value of the amplitude of the wave at $t=0s$ was given we could find the initial phase of the wave directly.
 $\bullet$ The equation of standing waves always has a fixed value of $x$ . The figure of the wave given here is a standing wave since, only the variation of the displacement of the wave with time is given at $x=8m$ , but it is not actually a standing wave since it is told that the wave is propagating along the positive X-axis.