Horizontal mass spring system is good but vertical mass spring system confuses me. Q1. Can there be two restoring forces in an SHM? Q2. If no, then weight of mass seems to disturb SHM as down extreme position below the mean position would be much farther than extreme position above mean position. If vertical mass spring system executes SHM, please elaborate how?mass spring system

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    $\begingroup$ The net force should still vary like it would in a horizontal spring. All gravity is really doing is changing the equilibrium position as far as I know. $\endgroup$
    – JMac
    Commented May 10, 2017 at 16:07

2 Answers 2


Gravity is just a constant force, so all it does is just shift the spring force linearly. This means all that happens is that everything gets shifted downward and that's it.

With a horizontal spring, we have $F = -kx$ as usual.

With a vertical spring, there are now two forces: the spring restoring force $F_{1} = -kx$, and gravity $F_{2} = mg$. You find the total force by adding both of the forces, and then you get $F = -kx + mg$.

Now note that this is the same as the horizontal case but with a constant added as a force. What this means is that we can take

$$F = -kx + mg = -k\left( x - \left( \tfrac{m}{k} \right)g \right)$$

by factoring. This equation is the same one as the one for the horizontal spring, but $x$ is just shifted by $x_{0} = \tfrac{mg}{k} $.

Can there be two restoring forces in an SHM?

Yes, there can be two restoring forces. Anytime you have two forces, you just add them together. If you add two restoring forces, you will still get SHM!

However, gravity is not a restoring force. It is just a constant downward force. Nonetheless, you still do what I did above and just add all of your forces together.


On applying a constant force in an SHM,it only changes its mean position BUT its amplitude,time-period, max velocity etc remains same.This can proved by solving the differential equation given below: $$\frac{d^2x}{dt^2} = \frac{-kx}{m}+g$$

(solution of above differential equation left to you)
On solving this differential equation,assuming mass to be at mean position ($\frac{d^2x}{dt^2} = 0$) at t=0; $$x=\frac{mg}{k}+Asin(\left(\frac{\sqrt k}{\sqrt m}t\right)$$ From above equation,it can be concluded that time period remains same while application of constant force in SHM


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