InterviewSolution
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InterviewSolution
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A particle moves with simple harmonic motion in a straight line. In first `taus`, after starting form rest it travels a destance a, and in next `tau s` it travels 2a, in same direction, then:A. amplitude of motion is 3aB. time period of oscillations is `8 pi`C. amplitude of motion is 4aD. time period of oscillations is `6 tau` |
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Answer» Correct Answer - D In SHM, a particle starts from rest, we have i.e. `x = A cos omega t, "at t"=0, x = A` When `t = tau`, then `x = A - a" "...(i)` When `t = 2 tau`, then `x = A - 3a" "...(ii)` On comparing Eqs. (i) and (ii), we get `A - a = A cos o omega tau` `A - 3a = A cos 2 omega tau` As, `cos 2 omega tau = 2 cos^(2)omega tau - 1` `rArr" "(A-3a)/(A)=2((A-a)/(A))^(2)-1` `= (2A^(2)+2a^(2)-4Aa-A^(2))/(A^(2))` `A^(2)-3aA=A^(2)+2a^(2)-4Aa` `rArr" "2a^(2)=aA` Now, `A - a = A cos omega tau" "[because tau = t]` `rArr" "cos omega t = 1//2 rArr (2pi)/(T)tau=(pi)/(3)rArr T = 6 tau` |
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