A simple harmonic motion of amplitude A has a time period T. The acceleration of the oscillator when its displacement is half the amplitude is

Options

A. `(4pi^(2)A)/(T^(2))`
B. `(2pi^(2)A)/(T^(2))`
C. `-(4pi^(2)A)/(T^(2))`
D. `-(2pi^(2)A)/(T^(2))`

Answer

D

Explanation

Acceleration `=-omega^(2)x=(-4pi^(2))/(T^(2))(A)/(2)=(-2pi^(2)A)/(T^(2))`