JEE Mains · Physics · STD 11 - 13. oscillations
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 \(\frac{{T}}{3}\)
- C \(\frac{{T}}{4}\)
- D \(\frac{{T}}{6}\)
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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