Amplitude of a particle of mass 0.1 kg executing SHM is 0.1 m. At the mean position its kinetic energy is `8xx10^(-3)J`. Find the time period of its vibration.

Explanation

At the mean position, the potential energy of the particle = 0
Hence, kinetic energy of the particle at the mean position = total energy = `1/2momega^(2)A^(2)`
`therefore" "1/2momega^(2)A^(2)=Eor,omega^(2)=(2E)/(mA^(2))or,omega=sqrt((2E)/(mA^(2)))`
i.e., time period
`T=(2pi)/omega=2pisqrt((mA^(2))/(2E))`
`=2xx3.14xxsqrt((0.1xx(0.1)^(2))/(2xx8xx10^(-3)))`
`=2xx3.14xxsqrt(1/16)=(2xx3.14)/4=1.57s`.