Two particles execute SHM 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 their amplitude. T

Question

Two particles execute SHM 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 their amplitude. The phase difference between them is:

TestHub · Accepted Answer

Let the displacement of the two particles be $y_{1} = A sin (ω t + ϕ_{1})$ and $y_{2} = A sin (ω t + ϕ_{2})$ .

They pass one another when $y_{1} = y_{2} = A /2$ and are going in opposite directions.

For $y = A /2$ , we have $sin (ω t + ϕ) = 1/2$ .

This implies $ω t + ϕ = \frac{π}{6}$ or $ω t + ϕ = \frac{5 π}{6}$ .

The velocity is $v = A ω cos (ω t + ϕ)$ . For opposite directions, the signs of their velocities must be opposite.

If one particle has phase $\frac{π}{6}$ , its velocity is positive.

If the other particle has phase $\frac{5 π}{6}$ , its velocity is negative.

The phase difference is $Δ ϕ = \frac{5 π}{6} - \frac{π}{6} = \frac{4 π}{6} = \frac{2 π}{3}$ .

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