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
Answer (1)
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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