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A particle executes SHM with a time period of `4 s`. Find the time taken by the particle to go directly from its mean position to half of its amplitude.A. 1sB. `(1)/(2)s`C. `(1)/(3)s`D. `(1)/(4)s` |
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Answer» Correct Answer - C `x=Asin(omegat+phi_(@))` at t=0, `x=0impliesAsinphi_(@)=0` or `phi_(@)=0` Hence `x=Asin(omegat)` or `A//2=Asin(omegat)` or 1/2`=sin(omegat)` `omegat=sin^(-1)((1)/(2))=(pi)/(6)` `t=(pi)/(6omega)=(pi.T)/(6(2pi))` as `omega=2pi//Timpliest T//12=1//3s` |
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