A body executes SHM such that its velocity at the mean positon is 1ms^(-1) and acceleation at the extreme position is 1.57ms^(-2). Calculate the amplitude and the time period of oscillation.

A body executes SHM such that its velocity at the mean positon is 1ms^

1.	A body executes SHM such that its velocity at the mean positon is 1ms^(-1) and acceleation at the extreme position is 1.57ms^(-2). Calculate the amplitude and the time period of oscillation.
Answer» Solution :The velocity at the MEAN POSITION is MAXIMUM given by `v_("MAX")=Aomega=1ms^(-1)` ……….1 The acceleration at the extreme position is maximum given by `a_("max")=Aomega^(2)=1.57ms^(-2)`……….2 Dividing 2 by 1 We get `(a_("max"))/(v_("max"))=(Aomega^(2))/(Aomega)=1.57/1` or `omega=1.57` i.e.`(2pi)/T=1.57` rad `:.` Time period `T=(2pi)/1.57=(2(3.14))/1.574s` From Eq. (1) `A omega =1` i.e `A(1.57)=1` or `A=1/1.57` `:.` Amplitude `A=0.637m`