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PhysicsSHMOscillations and SHMEasy2 min

PhysicsEasynumerical

A 0.2 kg particle is acted on by a restoring force $F = - 80 x$ N. Its speed at the mean position is 4 m/s. Find the amplitude, in m.

Answer:

0.20

Solution:

$ω = 80/0.2 = 20$ rad/s. $A = 4/20 = 0.2$ m.

Subject:PhysicsTopic:SHMSubtopic:Oscillations and SHM

⏱ 2m

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