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Physics Question on simple harmonic motion

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 AA and TT, respectively. At time t=0t = 0 one particle has displacement AA while the other one has displacement A2\frac{-A}{2} and they are moving towards each other. If they cross each other at time tt, then tt is :

A

T6\frac{T}{6}

B

5T6\frac{ 5T}{6}

C

T3\frac{T}{3}

D

T4\frac{T}{4}

Answer

T6\frac{T}{6}

Explanation

Solution

angle covered to meet =60=π3=60^{\circ}=\frac{\pi}{3} rad t=Qwt=\frac{Q}{w} =π3×2πT=T6=\frac{\pi}{3\times2\pi} T=\frac{T}{6}