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Q.A particle performing simple harmonic motion has an acceleration of 12 cm/s^2 at a distance of 3 cm from its mean position. Find its time period.

Rajasthan RbseRajasthan Board Senior Secondary Part-I Examination 2024Subjective· 2mImportance★★★★★
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Using a = ω^2 x with a = 12 cm/s^2 and x = 3 cm gives ω = 2 rad/s, so T = 2π/ω = π s ≈ 3.14 s.

In SHM, the magnitude of acceleration is related to displacement from the mean position by:

a = ω^2 x

Given a = 12 cm/s^2 and x = 3 cm:

ω^2 = a/x = 12/3 = 4

ω = 2 rad/s

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