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A particle executing simple harmonic motion of amplitude 5 cm has maximum speed of 31.4 cm/s. Its oscillation per second

1. 4
2. 3
3. 2
4. 1

Correct answer: 4

Solution

The maximum speed in SHM is given by vmax = Aω, where A is the amplitude and ω is the angular frequency. Substituting vmax = 31.4 cm/s and A = 5 cm, we get ω = vmax/A = 31.4/5 = 6.28 rad/s. The frequency f = ω/2π = 6.28/2π = 1 Hz. Therefore, the oscillation per second is 4.

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