Question:
A body is in simple harmonic motion with time period half second (T = 0.5 s) and amplitude one cm (A = 1 cm). Find the average velocity in the interval in which it moves from equilibrium position to half of its amplitude.
A body is in simple harmonic motion with time period half second (T = 0.5 s) and amplitude one cm (A = 1 cm). Find the average velocity in the interval in which it moves from equilibrium position to half of its amplitude.
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Average velocity = displacement / time; find the time from \(x = A\sin\omega t\) at \(x = A/2\).
Updated On: Nov 11, 2026
- 4 cm/s
- 6 cm/s
- 12 cm/s
- 16 cm/s
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The Correct Option is C
Solution and Explanation
Take \(x = A\sin\omega t\) starting from equilibrium. At \(x = A/2\): \(\sin\omega t = \frac12\), so \(\omega t = \frac\pi6\).
\(t = \frac{\pi/6}{2\pi/T} = \frac T{12} = \frac{0.5}{12} = \frac1{24}\) s.
Displacement = 0.5 cm.
Average velocity \(= \frac{0.5}{1/24} = 12\) cm/s.
Answer: (3) 12 cm/s
\(t = \frac{\pi/6}{2\pi/T} = \frac T{12} = \frac{0.5}{12} = \frac1{24}\) s.
Displacement = 0.5 cm.
Average velocity \(= \frac{0.5}{1/24} = 12\) cm/s.
Answer: (3) 12 cm/s
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