Wave Motion 5 Question 25
25. The displacement of the medium in a sound wave is given by the equation $y _i=A \cos (a x+b t)$ where $A, a$ and $b$ are positive constants. The wave is reflected by an obstacle situated a $x=0$. The intensity of the reflected wave is 0.64 times that of the incident wave.
(1991, 8M)
(a) What are the wavelength and frequency of incident wave?
(b) Write the equation for the reflected wave.
(c) In the resultant wave formed after reflection, find the maximum and minimum values of the particle speeds in the medium.
(d) Express the resultant wave as a superposition of a standing wave and a travelling wave. What are the positions of the antinodes of the standing wave? What is the direction of propagation of travelling wave?
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Answer:
Correct Answer: 25. (a) $\frac{2 \pi}{a}, \frac{b}{2 \pi}$
(b) $y_r=-0.8 A \cos (a x-b t)$
(c) $1.8 \mathrm{Ab}$, zero
(d) $y=-1.6 A \sin a x \sin b t+0.2 A \cos (a x+b t)$. Antinodes are at $x=\left[n+\frac{(-1)^n}{2}\right] \frac{\pi}{a}$. Travelling wave is propagating in negative $x$-direction.
Solution:
- (a) Wavelength of incident wave $=\frac{2 \pi}{a}$
and frequency of incident wave $=\frac{b}{2 \pi}$
(b) Intensity of reflected wave has become 0.64 times. But since $I \propto A^{2}$ amplitude of reflected wave will become 0.8 times.
$a$ and $b$ will remain as it is. But direction of velocity of wave will become opposite. Further there will be a phase change of $\pi$, as it is reflected by an obstacle (denser medium). Therefore, equation of reflected wave would be
$$ y _r=0.8 A \cos [a x-b t+\pi]=-0.8 A \cos (a x-b t) $$
(c) The equation of resultant wave will be,
$y=y _i+y _r=A \cos (a x+b t)-0.8 A \cos (a x-b t)$
Particle velocity
$v _p=\frac{\partial y}{\partial t}=-A b \sin (a x+b t)-0.8 A b \sin (a x-b t)$
Maximum particle speed can be $1.8 Ab$, where
$\sin (a x+b t)= \pm 1$ and $\sin (a x-b t)= \pm 1$
and minimum particle speed can be zero, where
$$ \sin (a x+b t) \text { and } \sin (a x-b t) \text { both are zero. } $$
(d) The resultant wave can be written as,
$$ \begin{aligned} y= & {[0.8 A \cos (a x+b t)-0.8 A \cos (a x-b t)] } \\ & +0.2 A \cos (a x+b t) \\ = & -1.6 A \sin a x \sin b t+0.2 A \cos (a x+b t) \end{aligned} $$
In this equation, $(-1.6 A \sin a x \sin b t)$ is the equation of standing wave and $0.2 A \cos (a x+b t)$ is the equation of travelling wave. The travelling wave is travelling in negative $x$-direction.
Antinodes are the points where, $\sin a x= \pm 1$
or
$$ a x=\left[n \pi+(-1)^{n} \frac{\pi}{2}\right] \text { or } x=\left[n+\frac{(-1)^{n}}{2}\right] \frac{\pi}{a} $$
