Question

Show that in S.H.M. the acceleration is directly proportional to its displacement at the given instant.

Answer 1

The acceleration of the particle executing SHM at any instant, is defined as the time rate of change of its velocity at that instant.

Acceleration, A = dv/dt = d/dt (aw cos ωt)

= -ω² a sin ωt    ....(i)

= -ω² y

At mean position, y = 0; from (i), A = 0 (min)

At extreme position, y = a; from (i) A - ω² a (max)

Thus, in S.H.M. the acceleration is also not uniform throughout the motion. It is minimum at the extreme position. The maximum value of acceleration is called acceleration amplitude in S.H.M.

In relation (i), negative sign shows that the acceleration (A) is directly opposite to the one in which displacement increases. Since, the displacement increases in the direction away from the mean position (on both the sides) acceleration is always directed towards the mean position in S.H.M. Also, A ∝ y, because ω is constant. Hence in S.H.M. the acceleration directly proportional to its displacement at the given instant.