Q. Find the period of a vibrating particle which has acceleration of 48 cm s - 2 , when displacement from mean position is 12 cm. Assume motion to be S.H.M.

a=0.48 m/s2sice accn is positive the displacement from the mean position is -ve that is -0.12 m angular position of the particle=sin--0.12Aa=-Aω2sinωt    accn eqn ωt=sin--0.12A   angular position for a=0.48 m/s20.48=-Aω2sinsin--0.12A0.48=ω20.12ω=4812=4=2 rad /sT=2πω=2π2=π sec time period Regards

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