InterviewSolution
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InterviewSolution
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The maximum speed of a particle performing a linear S.H.M. is 0.16 m/s and the maximum acceleration is 0.64 `m//s^(2)`. The period of S.H.M. is |
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Answer» Given : `v_(max)=0.16m//s`, `a_(max)=0.64m//s^(2)`, `T=?` We know that, Maximum velocity, `v_(max)=Aomega` Maximum acceleration `=Aomega^(2)` `:.omegaA=0.16` .....`(i)` `omega^(2)A=0.64` `omegaAxxomega=0.64` `0.16xxomega=0.64` [Using the value from equation `(i)`] `omega=4` Now, `omega=(2pi)/(T)` `(2pi)/(T)=4` `T=2pi//4` `=pi//2` `=1.57sec`. |
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