SATHEE: Wave Motion 5 Question 27

Wave Motion 5 Question 27

27. The following equations represent transverse waves;

$\begin{aligned} z_{1} = A \cos (k x - ω t); z_{2} = A \cos (k x + ω t) \\ z_{3} = A \cos (k y - ω t) \end{aligned}$

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)

Show Answer

Answer:

Correct Answer: 27. (a) $z_{1}$ and $z_{2}; x = (2 n + 1) \frac{π}{2 k}$ where $n = 0, \pm 1, \pm 2 \dots$ etc. (b) $z_{1}$ and $z_{3}, x - y = (2 n + 1) \frac{π}{k}$ where $n = 0, \pm 1, \pm 2 \dots$ etc.

Solution:

(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, $z = z_{1} + z_{2} = 2 A \cos k x \cos ω t = A_{x} \cos ω t$

Here, $A_{x} = 2 A \cos k x$

Resultant intensity will be zero, at the positions

where, $A_{x} = 0$

$\begin{array}{ll} or k x = (2 n + 1) \frac{π}{2} & where n = 0, \pm 1, \pm 2 \dots \dots . etc. \\ or x = (2 n + 1) \frac{π}{2 k} where n = 0, \pm 1, \pm 2 \dots \dots . etc. \end{array}$

(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

$z = z_{1} + z_{3} = 2 A \cos [\frac{k x + k y}{2} - ω t] \cos (\frac{k x - k y}{2})$

The resultant intensity is zero, where,

$\begin{aligned} \cos k (\frac{x - y}{2}) & = 0 \\ or & \frac{k (x - y)}{2} & = (2 n + 1) \frac{π}{2} \\ or & (x - y) & = (2 n + 1) \frac{π}{k} \end{aligned}$