In the case of simple harmonic motion of spring block system, why time period of the simple harmonic motion of the block is independent of acceleration of the system (spring-block system)?


1 Answer 1


A spring applies a force $F = kx$, where $k$ is the spring constant and $x$ is the distance from the rest position. If you have a mass $m$ on the spring, it will oscillate around $x = 0$ with period $t = 2\pi\sqrt{m\over{k}}$. If you then apply an acceleration $a$, the rest position of the spring will shift by $d_a = {a\over{k}}$. If you keep this acceleration constant, and displace the spring away from its new rest position $d_a$, then the net spring force will be $F_n = k(x - d_a)$. This is the same equation as you had before for the spring, just shifted by $d_a$. So, you will again get oscillation with period $t = 2\pi\sqrt{m\over{k}}$, except centered on $x = d_a$.

  • $\begingroup$ Why this shift causes no change in time period? Please explain. Thanks in advance $\endgroup$ Commented Oct 13, 2015 at 3:24
  • $\begingroup$ The oscillation period is a function of two things: the restoring force as a function of displacement from the spring's rest position, and the mass. Neither changes if you add an acceleration; you just shift the spring's rest position. $\endgroup$ Commented Oct 13, 2015 at 3:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.