When a particle performs a linear S.H.M. the ratio of its K.E. at the mean position to its P.E. at a point midway between the mean and extreme position is

Options

A. ` 4 : 1`
B. `1 : 4`
C. `3 : 2`
D. `2 : 3`

Answer

A

Explanation

K.E. at the mean position `(E_(1))=(1)/(2)m omega^(2)A^(2)`
P.E. at a dis"tan"ce, `x=(A)/(2)` is
`E_(2)=(1)/(2) m omega^(2)(x^(2))=(1)/(2)m omega^(2)(A^(2))/(4)`
`therefore (E_(1))/(E_(2))=(4)/(1)`