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JEE Main 2016 (9 Apr, Shift 1) · Periodic Motion & SHM Basics

Periodic Motion & SHM Basics question from JEE Main 2016 (9 Apr, Shift 1)

Q64JEE Main 2016 · 9 APR · SHIFT 1Medium

Periodic Motion & SHM Basics

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 −A2\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: (a)

  1. Take x=Asin⁡(ωt+ϕ)x = A\sin(\omega t + \phi). Particle 1 at x=Ax = A: ϕ1=π2\phi_1 = \frac{\pi}{2}.
  2. Particle 2 at −A2-\frac{A}{2} moving towards +x+x: ϕ2=−π6\phi_2 = -\frac{\pi}{6}.
  3. They cross when sin⁡(ωt+π2)=sin⁡(ωt−π6)\sin(\omega t + \frac{\pi}{2}) = \sin(\omega t - \frac{\pi}{6}), i.e. ωt+π2=π−ωt+π6\omega t + \frac{\pi}{2} = \pi - \omega t + \frac{\pi}{6}.
  4. 2ωt=2π3⇒ωt=π3⇒t=T62\omega t = \frac{2\pi}{3} \Rightarrow \omega t = \frac{\pi}{3} \Rightarrow t = \frac{T}{6}
  5. So the answer is (a) T6\frac{T}{6}.
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