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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/
Maximum velocity occurs at mean position in the simple harmonic motionK.E at the centre= 9-5 = 4J=>1/2mv^2 = 4 => V(max) =2m/sIn SHM V(max) = AW where A=1 in the given problem =>W=2 =>f=3.14 (f=2*3.14/W)
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)
Dear student,Maximum velocity occurs at mean position in the simple harmonic motionK.E at the centre= 9-5 = 4J=>1/2mv^2 = 4 => V(max) =2m/sIn 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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