A body of mass 25 gm performs linear S.H.M. The force constant of the motion is 400 dyne/cm. When the body is at a distance of 10 cm from the equilibrium position has velocity of 40 cm/s, then the total energy of the body will be

Options

A. `40xx10^(4)J`
B. `4xx10^(4)erg`
C. `2xx10^(4)erg`
D. `2xx10^(5)J`

Answer

B

Explanation

`omega=sqrt((k)/(m))=sqrt((400)/(25))=(20)/(5)=4`
`v=omegasqrt(A^(2)-x^(2))therefore 40=4sqrt(A^(2)-100)`
`100=A^(2)-100 therefore A^(2)=200`
`T.E.=(1)/(2)kA^(2)=(1)/(2)xx400xx200=4xx10^(4)erg`