“Equilibrium position of a massless spring vs its mean position”

I am confused about these two terms. Do they mean the same thing? I understand the term ‘mean position’. Let’s say one end of a vertical massless spring is attached to the ceiling and the other end is attached to a mass m. When the mass is released, the spring stretches and at one point, mg equals the spring force kx. And that’s the mean position, at x = $\frac{mg}{k}$

Is equilibrium position just another term for mean position? And since mean position equals $\frac{mg}{k}$, can I say that mean position of a spring is different for different masses?


1 Answer 1


The equilibrium position is the position of the mass when the system is stationary, in equilibrium.

The mean position is the mid point of the simple harmonic motion when the spring is oscillating.

Clearly for simple harmonic motion, which is symmetrical about the equilibrium position the two are equal. Only the situation to which the terms apply is different.

  • $\begingroup$ Thanks a lot. That cleared it up for me $\endgroup$
    – 4d_
    Apr 8, 2020 at 9:17

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.