Given, force F=(–25x)N
⇒0.25a=−25x
⇒a=−100x
Now comparing it with standard equation, a=−ω2x we get
ω=10.
Now, vmax=ωA
⇒ωA=4
⇒A=104=0.4m
⇒A=40cm.
A particle of mass 250g executes a simple harmonic motion under a periodic force F=(–25x)N. The particle attains a maximum speed of 4ms−1 during its oscillation. The amplitude of the motion is ______cm.
Held on 29 Jan 2023 · Verified 6 Jul 2026.
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