MHT CET · Physics · Oscillations
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 \(\left[\sin 30^{\circ}=0.5\right]\)
- A \(\frac{1}{4} \mathrm{~S}\)
- B \(\frac{3}{4} s\)
- C \(\frac{3}{2} s\)
- D \(\frac{1}{2} \mathrm{~s}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{4} \mathrm{~S}\)
Step-by-step Solution
Detailed explanation
(D)
T \(=3 \sec\)
\(y=A \sin \omega t\)
\(\frac{A}{2}=A \sin \frac{2 \pi}{T} t\)
\(\sin \frac{\pi}{6}=\sin \frac{2 \pi}{3} t\)
\(\frac{\pi}{6}=\frac{2 \pi}{3} t \quad \therefore t=\frac{\pi}{6} \times \frac{3}{2 \pi}=\frac{1}{4} s\)
T \(=3 \sec\)
\(y=A \sin \omega t\)
\(\frac{A}{2}=A \sin \frac{2 \pi}{T} t\)
\(\sin \frac{\pi}{6}=\sin \frac{2 \pi}{3} t\)
\(\frac{\pi}{6}=\frac{2 \pi}{3} t \quad \therefore t=\frac{\pi}{6} \times \frac{3}{2 \pi}=\frac{1}{4} s\)
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