A particle executes SHM represented by the equation =0.02sin (3.14t + (pi)/2) metre. Find (i) amplitude (ii) time period (iii) frequency (iv) epoch (v) maximum velocity and (vi) maximum acceleration.

A particle executes SHM represented by the equation =0.02sin (3.14t +

1.	A particle executes SHM represented by the equation =0.02sin (3.14t + (pi)/2) metre. Find (i) amplitude (ii) time period (iii) frequency (iv) epoch (v) maximum velocity and (vi) maximum acceleration.
Answer» Solution :COMPARE the equation `y=0.02sin (3.14+(pi)/2)` with general from of the equation `y=A sin (omegat+phi_(0))` (i) Amplitude : A=0.02m (ii) Time period (T) is given by `T=(2pi)/(omega)` or `T=(2pi)/3.14=2s` (iii) Frequency `v=1/T=1/2Hz=0.5Hz` (iv) Epoch `varphi_(0)=(pi)/2=3.14/2=1.57` rad (v) Maximum VELOCITY `v_("MAX")=Aomega=0.02xx3.14` `=0.68ms^(-1)` (vi) maximum acceleration `a_("max")=-Aomega^(2)=-0.02xx(3.14)^(2)=0.1972ms^(-2)`