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
Pick an option to check your answer.
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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