Question

A plane simple harmonic progressive wave of wavelength $120\text{cm}$ and speed $34800\text{cm/sec}$ is incident normally on a plane surface which is a perfect reflector of sound. Stationary waves are formed. The ratio of the amplitudes of vibration at points distant (i) $10\text{cm}$ and (ii) $30\text{cm}$ from the reflector is:

• A. $1:2$
• B. $1:0$
• C. $1:1$
• D. $1:4$

Answer 1

The correct option is A $1:2$

Equation of incident & reflected waves can be written as:
$y_i=a\sin \frac{2\pi }{\lambda }(vt-x)$
and $y_r=-a\sin \frac{2\pi }{\lambda }(vt+x)$
$\therefore y=y_i+y_r=-2a\sin \frac{2\pi x}{\lambda }\cos \frac{2\pi vt}{\lambda }$
is the equation of the stationary wave formed

Now, amplitude $A$ of the stationary wave is given by
$A=-2a\sin \frac{2\pi x}{\lambda }$
Here, $\lambda =120\text{cm}$

When $x=10\text{cm}$,
$A^{\prime }=-2a\sin 2\pi \left(\frac{10}{120}\right)=-2a\sin \frac{2\pi }{12}=-a$
When $x=30\text{cm}$,
$A^{\prime \prime }=-2a\sin 2\pi \left(\frac{30}{120}\right)=-2a$

$\therefore \frac{A^{\prime }}{A^{\prime \prime }}=\frac{-a}{-2a}=\frac{1}{2}$
Hence, the correct answer is option (a).

Why this question? When a progressive wave gets reflected at a fixed end, its amplitude and wave velocity both get reversed.