A particle performs simple harmonic motion with period of 3 second. The time taken by it to cover a distance…

A particle performs simple harmonic motion with period of 3 second. The time taken by it to cover a distance equal to half the amplitude from mean position is $\left[\sin 30^{\circ}=0.5\right]$
  1. $\frac{1}{4} \mathrm{~S}$
  2. $\frac{3}{4} s$
  3. $\frac{3}{2} s$
  4. $\frac{1}{2} \mathrm{~s}$

Solution

T $=3 \sec$ $y=A \sin \omega t$ $\frac{A}{2}=A \sin \frac{2 \pi}{T} t$ $\sin \frac{\pi}{6}=\sin \frac{2 \pi}{3} t$ $\frac{\pi}{6}=\frac{2 \pi}{3} t \quad \therefore t=\frac{\pi}{6} \times \frac{3}{2 \pi}=\frac{1}{4} s$

Asked in: MHT CET 2020 (16 Oct Shift 2)

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