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InterviewSolution
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A particle of mass 0.2 kg is excuting SHM of amplitude 0.2 m. When it passes through the mean position its kinetic energy is `64 xx 10^(-3) J` . Obtain the equation of motion of this particle if the initial phase of oscillation is `pi//4`. |
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Answer» Here, A=0.2 m `phi =(pi)/(4)` As we know , KE`=(1)/(2)momega^(2)A^(2)` where , m=mass of particle =0.2 kg `64 xx10^(-3) =(1)/(2)xx0.2xx omega^(2)xx0.2^(2)` `omega^(2)=(128 xx10^(-3))/(0.2xx0.2xx0.2) "or" omega=4s^(-1)` `therefore` Equation of motion can be written as `x=A"sin"(omegat+phi)` `=0.2 "sin"(4t+(pi)/(4))` |
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