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 $t=\frac{T}{6}(T$ is the time period$)$. Assume that the source was at mean position at $t=0$. The wavelength of the running wave is

• A. $0.96m$
• B. $0.48m$
• C. $0.24m$
• D. $0.12m$

Answer 1

The correct option is B $0.48m$

Going by the data given to us, this wave is sinusoidal in nature and the wave equation takes the form of
$y=A\sin (\omega t-kx),$ as it is given that at $t=0,$ the source is at mean position.
Here$,x=4cm=0.04m$
$y=A/2$
Amplitude $=A$
$t=\frac{T}{6}$
We know that $\omega =2\frac{\pi }{T}$
$\Rightarrow \frac{A}{2}=A\sin \left(\right(2\pi /T\left)\right(T/6)-0.04k)$
$\sin \left(\right(2\pi /T\left)\right(T/6)-0.04k)=1/2$
$\Rightarrow \left(\right(2\pi /T\left)\right(T/6)-0.04k)=\pi /6$
$k=\pi /0.24$
wavelength $\lambda =2\pi /k=0.48m$