A body executes SHM, such that its velocity at the mean position 1ms^(-1) and acceleration 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 position 1ms^(

1.	A body executes SHM, such that its velocity at the mean position 1ms^(-1) and acceleration 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 exreme position is maximum given by, `a_("max")= Aomega^(2)=1.57 ms^(-2)` ………(2) Dividing (2) by (1) we get `(a_("max"))/(v_("max"))= (Aomega^(2))/(Aomega)= (1.57)/(1)` `:. ` TIME PERIOD `T= (2pi)/(1.57)= (2(3.14))/(1.57)4s` From Eq. (1) `Aomega=1` i.e., `A(1.57)= 1` or `A= (1)/(1.57)` `:.` Amplitude A= 0.637 m