Question

A point source is producing sinusoidal transverse waves. If the displacement from the equilibrium position of a point $4\text{cm}$ from source is half the amplitude of wave at $t=\frac{T}{6}$ ($T$ being time period). The wavelength of travelling wave is

• A. $0.48\text{m}$
• B. $0.96\text{m}$
• C. $0.24\text{m}$
• D. $0.12\text{m}$

Answer 1

The correct option is A $0.48\text{m}$

For the source(s) at mean position at t = 0, the equation of transverse wave produced will be,
$y=A\sin (\omega t-kx)$
here $x=4\text{cm}$ (distance of point from source)


at $t=\frac{T}{6},y=\frac{A}{2}$
$\Rightarrow \frac{A}{2}=A\sin [(\frac{2\pi }{T}\times \frac{T}{6})-(\frac{2\pi }{\lambda }\times 4)]$
$\Rightarrow \frac{\pi }{3}-\frac{8\pi }{\lambda }=\frac{\pi }{6}$
$[\because {\sin }^{-1}\left(\frac{1}{2}\right)=\frac{\pi }{6}]$
$\therefore \lambda =48\text{cm}=0.48\text{m}$