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A particle executes \( \boldsymbol{S} \boldsymbol{H} \boldsymbol{M} \) of amplitude \( 25 c m \) and time period \( 3 s \) What is the minimum time required for the particle to move between two points \( 12.5 \mathrm{cm} \) on either side of the mean position?

1. \( 0.5 s \)
2. \( 1.0 s \)
3. \( 1.5 s \)
4. 2.0

Correct answer: \( 0.5 s \)

Solution

For SHM, the minimum time between two symmetric positions about the mean is found from the phase difference, not distance. Here 12.5 cm is half the amplitude, so the particle moves from one side to the other in a quarter of a period: T/4 = 3/4 = 0.75 s? Wait—minimum time means from +12.5 cm to -12.5 cm along the shortest phase path, which is half a period only if starting at an extreme; but for these positions the phase difference is 60°, giving T/6 = 0.5 s.

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