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Grade 12Mechanics

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/

Question image for The potential energy of a harmonic oscillator of m
Profile image of Tejas
8 Years agoGrade 12
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2 Answers

Profile image of Arun
8 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)

Profile image of Kushagra Madhukar
6 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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