# 8.2.1 Restoring Force

if we multiply the mass of the oscillating body m to both sides of the a-x equation, we get $\displaystyle \displaystyle ma=-m{{\omega }^{2}}x$

But $\displaystyle {{F}_{{net}}}=ma$, so $\displaystyle \displaystyle {{F}_{{net}}}=-m{{\omega }^{2}}x$

This tells us that a SHM experiences a net force that is (1) proportional to its displacement and (2) opposite in direction to its displacement. Since this net force is always directed towards the equilibrium position, it is called the restoring force.

Take for example the spring-mass system.

(a) $\displaystyle x=0$

At the equilibrium position, the downward weight mg is balanced by the upward tension force T of the spring. The restoring force is zero at the equilibrium position.

(b) $\displaystyle x>0$(We are using the “downward is positive” sign convention)

When the mass is below the equilibrium position, so the spring is further extended by x (compared to when it is at the equilibrium position). This results in the upward tension increasing by kx. But the downward weight is fixed as mg. So the net force is simply kx and upward.

(c) $\displaystyle x<0$(We are using the “downward is positive” sign convention)

When the mass is below the equilibrium position, so the spring is shorter by x (compared to when it is at the equilibrium position). This results in the upward tension decreasing by kx. But the downward weight is fixed as mg. So the net force is simply kx and downward.

In summary, the magnitude of the net force is kx, and its direction is always towards the equilibrium position. So the spring-mass system does indeed have a restoring force whose magnitude is proportional to the displacement, directed towards the equilibrium position. This is why when the mass is displaced from the equilibrium position and released, it goes into simple harmonic motion.

Concept Test

1422