How much displacement will a coil spring with a spring constant of $120N{{m}^{-1}}$ achieve if it is stretched by a $60N$ force?

Hint: When the spring is stretched, a restoring force is developed in it which is equal to the external force but acts in the opposite direction. Since, the spring follows harmonic motion, the force is a product of spring constant and its displacement. Substituting corresponding values in the above relation, we can calculate the displacement.
Formulas used:
$F=-kx$

Complete answer:
The restoring force developed in a spring when it is stretched is directly proportional to the negative of its displacement; hence the spring follows harmonic motion. Therefore,
$F\propto -x$
Here, $F$ is the force
$x$ is the displacement
On removing the sign of proportionality,
$F=-kx$
Here, $k$ is the spring constant and is defined as the force per unit displacement.
Give, force applied on the spring is $60N$, the spring constant is $120N{{m}^{-1}}$
In the above equation, substituting given values to get,
$\begin{align}
  & 60=120\times x \\
 & \Rightarrow x=\dfrac{60}{120} \\
 & \Rightarrow x=\dfrac{1}{2} \\
 & \therefore x=0.5m \\
\end{align}$
The spring is stretched by $0.5m$.
Therefore, the displacement of the spring is $0.5m$.

Additional Information:
Periodic motion is that motion which repeats itself after a certain interval. Some examples of periodic motion are rotation, revolution, harmonic motion. In Harmonic motion, a body travels between its mean and extreme positions such that the restoring force is directly proportional to the negative of displacement. Its direction is towards the equilibrium position.

Note:
The restoring force is equal to the external force. Restoring force is responsible for bringing the spring back to its equilibrium position and is developed internally. The restoring force acts opposite to the displacement, this is indicated by the negative sign. The potential energy is maximum at extreme positions while the kinetic energy is highest at the mean position.