The kinetic energy of a particle executing SHM is 16J. When it is in its mean position. If the amplitude of oscillation is 25cm and the mass of the particle is 5.12 kg, the time period of its oscillation in second is

Options

A. `(pi)/(3) sec`
B. `(pi)/(5) sec`
C. `(pi)/(2) sec`
D. `(pi)/(10) sec`

Answer

B

Explanation

In the mean position `K.E.=(1)/(2)m omega^(2)A^(2)`
`A=(1)/(4)m and m=5.12 kg`
`therefore 16=(1)/(2)xx5.12xx omega^(2)xx(1)/(4)xx(1)/(4)`
`therefore omega^(2)=(512)/(5.12)=100`
` therefore omega =(2pi)/(T)=10 therefore T=(2pi)/(10)=(pi)/(5) sec.`