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MHT CET 2025 (25 Apr, Shift 1) · Periodic Motion & SHM Basics

Periodic Motion & SHM Basics question from MHT CET 2025 (25 Apr, Shift 1)

Q21MHT CET 2025 · 25 APR · SHIFT 1Medium

Periodic Motion & SHM Basics

Two particles ' AA ' and ' BB ' execute SHMs of periods ' TT ' and 3T2\frac{3T}{2}. If they start from the mean position then the phase difference between them, when the particle ' AA ' completes two oscillations will be

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Answer: (c)

  1. Phase gained =2πtT= \frac{2\pi t}{T}. A completes 2 oscillations, so t=2Tt = 2T.
  2. ϕA=2π(2)=4π\phi_A = 2\pi(2) = 4\pi, ϕB=2π(2T)3T/2=8π3\phi_B = \frac{2\pi(2T)}{3T/2} = \frac{8\pi}{3}
  3. Δϕ=4π−8π3=4π3\Delta\phi = 4\pi - \frac{8\pi}{3} = \frac{4\pi}{3}
  4. So the answer is (c) 4π3\frac{4\pi}{3}.
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