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If the displacement of a moving point at any time is givenby an equation of the form `y (t) = a cos omega t + b sin omega t`, shown that the motion is simple harmonic . If `a = 3 m, b = 4m and omega = 2`: determine the period , amplitude, maximum velocity and maximum acceleration. |
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Answer» `y=acos omegat+bsin omegat` Let, `a=Rsin theta and b=Rcos theta` Then, `y=Rsintheta cos omega t +R cos theta sin omega t` `y=Rsin (omega t + theta)` Velocity, `V=(dy)/(dt)=Romega cos (omegat+theta)` Acceleration, `A=(dV)/(dt)=-omega^(2)Rsin(omegat+theta)` `=omega^(2)y` It means `A prop y` and `-ve` sign shows that A is directed towards the mean position, hence the motion is SHM. Amplitude, `R=sqrt(a^(2)+b^(2))=sqrt(3^(2)+4^(2))` `=5` units Time period, `T=(2pi)/(omega)=(2pi)/(2)=3.14s` Max. velocity, `V_(max)=Romega=5xx2=10unit` Max. acceleration `=-omega^(2)R` `=-(2)^(2)xx5=-20units` |
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