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:

• A. $\frac{5T}{6}$
• B. $\frac{T}{4}$
• C. $\frac{T}{3}$
• D. $\frac{T}{6}$

Answer 1

Answer: (D)

$\frac{T}{6}$

Equations of two particles can be written as $x_1=Acos\left(\omega t\right)$ and $x_2=Asin(\omega t-\frac{\pi }{6})$.

Equating the two gives $t=\frac{T}{6}$.