At position x from the mean position in S.H.M.
velocity, v=ωA2−x2
acceleration, a=xω2
here ω is the angular frequency and A is amplitude.
Given in the problem, v=a
⇒ ωA2−x2=xω2
⇒ (3)2−(2)2=2(T2π)
5=T4π
Time period, T=54π
A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is
Held on 30 Apr 2017 · Verified 9 Jul 2026.
π5
2π5
54π
32π
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