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If the maximum acceleration of a particle performing S.H.M. is numerically equal to twice the maximum velocity then the period will beA. 1.57 sB. 3.142 sC. 6.28 sD. 2 s |
Answer» Correct Answer - B `a_(m)=2v_(m)` `Aomega^(2)=2Aomega` `omega=2` `T=(2pi)/(omega)=(2pi)/(2)=pi=3.142 s` |
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