Question

Two particles executes s.H.M. Of same amplitude and frequency along the same straight line. They pass one another when going in opposite directions. Each time their displacement is half of their amplitude. The phase difference between them is

Answer 1

Answer: the phase difference = $\dfrac{4\pi}{6}$

Explanation: we know that,

The wave equation is

$x = A sin (\omega t +\delta)$

Where, x = displacement

A = amplitude

$\delta$= phase constant

Given that, at each time the displacement is half of their amplitude

So, for first particle

At $x = \dfrac{A}{2}$ and t = 0

Now, $\dfrac{A}{2} = A sin \delta$

$sin\delta = \dfrac{1}{2}$

$\delta = \dfrac{\pi}{6}, \dfrac{5\pi}{6}$

Now, the velocity is

$v = \dfrac{dx}{dt}$

$v = A\omega cos(\omega t+\delta)$

Now, at t= 0, and $\delta = \dfrac{\pi}{6}$

The velocity is

$v = A\omega cos\dfrac{\pi}{6}$

$v = A\omega \dfrac{\sqrt3}{2}$

The velocity is positive for first particle

Now, at t= 0, and $\delta = \dfrac{5\pi}{6}$

$v = -A \omega\dfrac{\sqrt3}{2}$

The velocity is negative for second particle.

Now, the phase difference between them

The phase difference $= \dfrac{\pi}{6} - \dfrac{5\pi}{6}$

So, the phase difference = $\dfrac{4\pi}{6}$.