Question

The minimum distance between two particles in same phase is $18cm$. If the velocity of the wave is $36m{s}^{-1}$, then find the time interval after which the given particle undergoes a phase change of $\pi$.

Answer 1

The minimum distance between two particles in same phase is the wavelength $\lambda$

$\therefore \lambda =18cm=18\times {10}^{-2}$ metres

velocity $v=36m/s$

Time to complete one oscillation is given by $T=\frac{\lambda }{v}$

The time interval after which the given particle undergoes a phase change of $\pi$, $T^{\prime }$ is the half of time period.

$T^{\prime }=\frac{\lambda }{2v}=\frac{18\times {10}^{-2}}{2\times 36}=0.25sec$