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A particle starts simple harmonic motion from the mean position. Its amplitude is A and time period is T. What is its displacement when its speed is half of its maximum speed?

1. √2/3 A
2. √3/2 A
3. 2/√3 A
4. A/√2

Correct answer: 2/√3 A

Solution

The maximum speed in SHM is given by vmax = ωA, where ω = 2π/T. When the speed is half of vmax, we use the SHM velocity equation v = ω√(A² - x²). Solving for x when v = vmax/2 gives x = 2A/√3.

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