A particle of mass `m` executes `SHM` with amplitude 'a' and frequency 'v'. The average kinetic energy during motion from the position of equilibrium to the end is:

Options

A. `pi^(2)ma^(2)v^(2)`
B. `1/4 ma^(2)v^(2)`
C. `4pi^(2)ma^(2)v^(2)`
D. `2pi^(2)ma^(2)v^(2)`

Answer

A

Explanation

Mass `=` m, amplitude `=a`, frequnecy `= v`
`(KE)_(epsiV) = (1)/(4)m(2piv^(2))a^(2) = pi^(2)mv^(2)a^(2)`