Answer to Question #112396 in Classical Mechanics for Pravar Gupta

Question

If a mass 'm' is attached to a spring and the mass is brought down slowly such that its spring force balances the gravity then the extension will be (x = mg/k) and when the spring is allowed to free fall then the extension is (x = 2mg/k). So could you please tell me in which condition stable equilibrium is acheived and in which unstable equilibrium??

Answer 1

As you correctly mentioned, stable equilibrium is achieved when the mass is brought down slowly. Then the extension will be x=mg/k if the mass does not oscillate. In the real world the mass on the spring will achieve stable equilibrium in all experiments because such oscillations are damped, that is, they decrease the amplitude with every oscillation if we do not apply any external forces other than gravity.

On the whole the concept of equilibrium equilibrium can be understood with the following figure:

Therefore, in the real world, the mass hanging on a spring will represent stable equilibrium: we can displace the resting mass and it will eventually stop oscillating again.

In the ideal world, the mass hanging on a spring will represent unstable equilibrium: if the mass is in equilibrium and we displace it, it will start oscillations and will never stop.