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JEE Mains · Physics · STD 11 - Oscillations
  • JEE Mains 2016
  • Hard
  • MCQ

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

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Answer & Solution

Correct Answer

(D) \(\frac{{T}}{6}\)

Step-by-step Solution

Detailed explanation

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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