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InterviewSolution
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The time period of a particle in simple harmonic motion is T. Assume potential energy at mean position to be zero. After a time of `(T)/(6)` it passes its mean position its,A. velocity will be half its maximum velocityB. displacement will be half its amplitudeC. acceleration will be nearly `86%` of its maximumD. `KE=PE` |
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Answer» Correct Answer - A::C `y=asinomegat=asin((2pit)/(T))` `v=(dy)/(dt)=omegaacos((2pit)/(T))` At `t=(T)/(6)`,`v=omegaacos(((2pi)/(T)(T)/(6)))=(1)/(2)omegaa` or, `v=((v_(max))/(2))` `y=asin((2pi)/(T))xx(T)/(6)=asin(pi)/(3)` It is not half of a. acceleration `=(d^2y)/(dt^2)=(dv)/(dt)=omega^2asin((2pit)/(T))` `=omega^2asin((pi)/(3))=0.86(AC)_(max)` At this instant, `KE=(1)/(2)mv^2=(1)/(2)m((v_(max))/(2))^2=((KE)_(max))/(4)=(1)/(4)(TE)` `PE=TE-KE=(3)/(4)(TE)` i.e., `KEnePE` |
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