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A linear harmonic oscillator of force constant `2 xx 10^6 N//m` and amplitude (0.01 m) has a total mechanical energy of (160 J). Its. |
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Answer» Here, `k=2xx10^(6)Nm^(-1)`, `a=2cm =0.02m` ltbtgt Total energy `=600J` As maximum K.E. `=`max. P.E. `=(1)/(2)ka^(2)` `=(1)/(2)xx(2xx10^(6))xx(0.02)^(2)=400J` In SHM P.E. is minimum at the mean position and K.E. is maximum at the mean position, whereas total meachanical energy is constant throughout. Therefore, at mean position total mechnaical energy `=max K.E.+min. P.E.` or `600=400+min. P.E.` or min. P.E.` =600-400=200J` |
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