Question

A particle starts from mean position and performs S.H.M. with period 6 second. At what time its kinetic energy is 50% of total energy?

`(cos45^circ=1/sqrt2)`

Options

• 1 second

• 0.50 second

• 0.25 second

• 0.75 second

Answer 1

0.75 second

Explanation:

T = 6 sec

`1/2momega^2(A^2-x^2)=1/2(1/2momega^2A^2)`

`A^2-x^2=A^2/2`

`thereforex^2=A^2/2`

`=>x=A/sqrt2`

`A/sqrt2=Asinomegat=Asin  (2pi)/Tt`

`1/sqrt2=sin(2pi)/6t=sin  pi/3t`

`sin  pi/4t=sin  pi/3t`

`thereforet=3/4=0.75sec`