Wave Motion Ques 125
- The following equations represent transverse waves;
$$ \begin{aligned} & z _1=A \cos (k x-\omega t) ; z _2=A \cos (k x+\omega t) \\ & z _3=A \cos (k y-\omega t) \end{aligned} $$
Identify the combination (s) of the waves which will produce (a) standing wave (s), (b) a wave travelling in the direction making an angle of 45 degrees with the positive $X$ and positive $Y$-axes. In each case, find the position at which the resultant intensity is always zero.
$(1987,7 M)$
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Answer:
Correct Answer: 125.(a) $z _1$ and $z _2 ; x=(2 n+1) \frac{\pi}{2 k}$ where $n=0, \pm 1, \pm 2 \ldots$ etc.
(b) $z _1$ and $z _3, x-y=(2 n+1) \frac{\pi}{k}$ where $n=0, \pm 1, \pm 2 \ldots$ etc.
Solution:
Formula:
- (a) For two waves to form a standing wave, they must be identical and should move in opposite directions. Therefore, $z _1$ and $z _2$ will produce a standing wave. The equation of standing wave in this case would be, $z=z _1+z _2=2 A \cos k x \cos \omega t=A _x \cos \omega t$
Here, $\quad A _x=2 A \cos k x$
Resultant intensity will be zero, at the positions
where, $A _x=0$
$$ \begin{array}{ll} \text { or } \quad k x=(2 n+1) \frac{\pi}{2} \quad \text { where } n=0, \pm 1, \pm 2 \ldots \ldots . \text { etc. } \\ \text { or } \quad x=(2 n+1) \frac{\pi}{2 k} \quad \text { where } n=0, \pm 1, \pm 2 \ldots \ldots . \text { etc. } \end{array} $$
(b) $z _1$ is a wave travelling in positive $X$-axis and $z _3$ is a wave travelling in positive $Y$-axis.
So, by their superposition a wave will be formed which will travel in positive $x$ and positive $y$-axis. The equation of wave would be
$$ z=z _1+z _3=2 A \cos \left[\frac{k x+k y}{2}-\omega t\right] \cos \left(\frac{k x-k y}{2}\right) $$
The resultant intensity is zero, where,
$$ \begin{aligned} & \cos k\left(\frac{x-y}{2}\right) & =0 \\ \text { or } & \frac{k(x-y)}{2} & =(2 n+1) \frac{\pi}{2} \\ \text { or } & (x-y) & =(2 n+1) \frac{\pi}{k} \end{aligned} $$
where $n=0, \pm 1, \pm 2, \ldots$ etc.
BETA
