Two particles are performing simple harmonic motion in a straight… - Vidyadip

Question

Two particles are performing simple harmonic motion in a straight line about the same equilibrium point. The amplitude and time period for both particles are same and equal to $A$ and $T,$ respectively. At time $t=0$ one particle has displacement $A$ while the other one has displacement $\frac {-A}{2}$ and they are moving towards each other. If they cross each other at time $t,$ then $t$ is

✓

Answer

Angle covered to meet $\theta=60^{\circ}=\frac{\pi}{3} \mathrm{rad}$

If they cross each other at time $t$ then

$t=\frac{\theta}{2 \pi}=\frac{\pi}{3 \times 2 \pi} \mathrm{T}=\frac{\mathrm{T}}{6}$

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