Question

A plane simple harmonic progressive wave given by the equation, $y=A\sin (\omega t-kx)$ of wavelength $120\text{cm}$ is incident normally on a plane surface which is a perfect reflector (acts as fixed end). If a stationary wave is formed, then the ratio of amplitudes of vibrations at points $10\text{cm}$ and $30\text{cm}$ from the reflector is

(Here $x$ is measured from reflector)

• A. $1:2$
• B. $1:3$
• C. $2:1$
• D. $3:1$

Answer 1

The correct option is A $1:2$

We know that, standard equation of standing wave is,
$y=2A\sin \left(kx\right)\cos \left(\omega t\right)$
At any particular instant i.e., $t=$ fixed, amplitude of particle is given by $A_0=2A\sin \left(kx\right)$
Wavenumber, $k=\frac{2\pi }{\lambda }=\frac{2\pi }{120}=\frac{\pi }{60}{\text{cm}}^{-1}$
So,
$A_0=2A\sin \left(\frac{\pi x}{60}\right)\cos \left(\omega t\right)$
At, $x=10\text{cm}$
$A_1=2A\sin \left(\frac{\pi \times 10}{60}\right)\cos \left(\omega t\right)=A\cos \left(\omega t\right)$
Similarly, at, $x=30\text{cm}$
$A_2=2A\sin \left(\frac{\pi \times 30}{60}\right)\cos \left(\omega t\right)=2A\cos \left(\omega t\right)$
Therefore,
$A_1:A_2=1:2$