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The potential energy of a harmonic oscillator of mass 2 kg in its mean position is 5 J. If its total energy is 9 J and its amplitude is 0.01 m, its time period will beI have no understood that why have they found out KE and the equated it to its formula Where as we have got formula for TE and we even know it`s formula Pls.explain This is the link of its soultion I got on your website http://www.askiitians.com/iit-study-material/iit-jee-physics/oscillation/energy-in-simple-harmonic-motion/

The potential energy of a harmonic oscillator of mass 2 kg in its mean position is 5 J. If its total energy is 9 J and its amplitude is 0.01 m, its time period will beI have no understood that why have they found out KE and the equated it to its formula Where as we have got formula for TE and we even know it`s formula Pls.explain This is the link of its soultion I got on your website http://www.askiitians.com/iit-study-material/iit-jee-physics/oscillation/energy-in-simple-harmonic-motion/

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Grade:12

2 Answers

Arun
25750 Points
6 years ago

Maximum velocity occurs at mean position in the simple harmonic motion

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)

Kushagra Madhukar
askIITians Faculty 628 Points
3 years ago
Dear student,
Maximum velocity occurs at mean position in the simple harmonic motion
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)
 
Thanks and regards,
Kushagra

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