A body of mass `m` is attached by an inelastic string to a suspended spring of spring constant `k`. Both the string and the spring have negligible mass and the string is inextensible and of length `L`. Initially, the mass `m` is at rest.
If the mass `m`, from the initial position of rest is pulled down a distance `A` and then released, assuming that the string remains taut throughout the motion, the maximum (downwards) acceleration of the oscillating body will be

Options

A. `(kA)/(m)`
B. `(kA)/(2m)`
C. `(g)/(2)`
D. `g`

Answer

A

Explanation

If the mass is pulled down by distance a from its equilibrium position and then released, it will perform SHM of amplitude A. Assuming that the string remains taut, the maximum acceleration will be at the amplitude point and given `f_m=Aomega^2`