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total energy of a harmonic oscillator of mass 2 kg is 9 joules. if its energy at rest is 5 its kinetic energy at the mean position will be


shashank kumar 29th Jan, 2019
Answer (1)
Debosmitha Bhattacharyya Student Expert 29th Jan, 2019

Hello Shashank,

Maximum velocity occurs at the mean position in the simple harmonic motion -

So K.E at the centre= 9-5 = 4J

=>1/2mv^2 = 4 => V(max) =2m/s

In SHM V(max) = AW where A=1 in the given problem =>W=2 =>f=3.14  (f=2*3.14/W)

Hope this helps.

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