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A linear harmonic oscillator of force ocnstant `2xx10^(6)N//m` and amplitude 0.01 m has a total mechanical energy of 160J. What is the maximum K.E. and minimum P.E.? |
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Answer» Here, maximum K.E. `=` max. PE `=(1)/(2)kr^(2)` `=(1)/(2)xx(2xx10^(6))xx(0.01)^(2)=100J` In SHM, P.E. is minimum at mean position and K.E. is maximum at mean position. But the total meachanical energy in SHM is constant. Therefore, at mean position., total mechanical energy `=` max. K.E. `+` minimum P.E. `160=100+` minimum P.E. So, minimum P.E. `=160-100=60J` |
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