Question

The displacement from the position of equilibrium of a point $4cm$ from a source of sinusoidal oscillations is half the amplitude at the moment $t=\frac{T}{6}\left(T\text{~is the time period}\right)$. Assume that the source was at mean position at $t=0$. The wavelength $\left(\text{in}m\right)$ of the running wave is

Answer 1

Given: $y=\frac{A}{2}m,t=\frac{T}{6}s,x=4cm$
Equation of wave is
$y=A\sin (\omega t-kx)$
$\Rightarrow \frac{A}{2}=A\sin (\frac{2\pi }{T}\times \frac{T}{6}-\frac{2\pi }{\lambda }\times 4)$
$\Rightarrow \frac{\pi }{6}=\frac{\pi }{3}-\frac{2\pi }{\lambda }\times 4$
$\Rightarrow \frac{2\pi }{\lambda }\times 4=\frac{\pi }{6}$
$\lambda =48cm$
$\lambda =0.48m$
Final answer: $\left(0.48\right)$