A particle of mass 10 g is performing SHM. Its kinetic energies are 4.7 J and 4.6 J when the displacements are 4 cm and 6 cm , respectively. Compute the period of oscillation.

Explanation

Data : m = 0.01 kg, `KE_(1) = 4.7 J, x_(1) = 4 xx 10^(-2) m, KE_(2) = 4.6 J `
` x_(2) = 6 xx 10^(-2) m`
Since the total energy of a particle is SHM is constant .
` KE_(1) + PE_(1) = KE_(2) +PE_(2)`
` KE_(1) -KE_(2) = PE_(2) -PE_(1) = 1/2m omega^(2) (x_(2)^(2) -x_(1)^(2))`
` 1/2 m((4pi^(2))/(T^(2)) (x_(2)^(2) -x_(1)^(2)) " " (therefore omega = (2pi)/T)`
` T^(2) = 2mpi^(2) = ((x_(2)^(2)-x_(1)^(2)))/(KE_(1) -KE_(2))`
` = 2(0.01) pi^(2) ([(6^(2)-4^(2)) xx 10^(-4)])/(4.7 - 4.6)`
` = 0.2 pi^(2) xx 20 xx 10^(-4) = 4pi^(2) xx 10^(-4)`
The period of oscillation of the particle.
` T= sqrt(4pi^(2) xx 10^(-4)) = 2pi xx10^(-2) = 6.284 xx 10^(-2)S`