According to Hooke’s law, \[\text{Restoring force (F)} = -kl\text{ ___(1)}\] where, $k =$ **force constant** or **spring factor** $k$ is defined as the restoring force per unit extension. There is a negative sign which shows that the restoring force acts in opposite direction i.e. upward. At this condition, there is no acceleration. It is because the restoring force is equal to the weight of the mass.

\[∴-kl=-mg\] \[kl=mg\] If the mass is displaced through a distance $y$ below the equilibrium as shown in figure (c), then the total extension becomes $(l+y)$ and according to the Hooke’s law, \[\text{New restoring force (F’)} = -k(l+y)\text{ ___(2)}\] subtracting $(1)$ from $(2)$, \[F’-F=-k(l+y)+kl\] \[F’-F=-ky\]

This force $F’-F$ produces oscillations in the spring. Then, the acceleration $a$ is given by, \[a=\frac{F’-F}{m}\] \[∴a=\frac{-ky}{m}\]

Here, $k$ and $m$ are constant. So, acceleration is directly proportional to the displacement from the mean position and is directed towards the mean position. Thus, the motion of a helical spring is Simple Harmonic Motion.

Then, Time period of the spring $(T)$, \[T=2π\sqrt{\frac{y}{a}}\] and, in magnitude only, \[a=\frac{k}{m}y\] \[∴T=2π \sqrt{\frac{y}{\frac{k}{m}y}}\] \[T=2π \sqrt{\frac{m}{k}}\] This gives the time period of the spring.

Consider a mass $m$ attached to massless spring fixed to a rigid support as shown in figure (a).

When this spring is stretched by a distance $y$ as shown in figure (b), the system develops a restoring force $(F)$ which tries to regain its original position.

According to Hooke’s law, \[F=-ky\text{ ___(1)}\] When the spring is released, due to the restoring force, the mass moves back to its original position. But due to its kinetic energy, it does not stop at its original position and undergoes compression as shown in figure (c). When the spring is compressed, the restoring force $F$ acts opposite to the motion of the mass.

If the mass gets an acceleration $a$, then \[F=ma\text{ ___(2)}\] From $(1)$ and $(2)$, \[-ky=ma\] \[∴a=-\frac{ky}{m}\]

Here, $k$ and $m$ are constant. So, acceleration is directly proportional to the displacement from the mean position and is directed towards the mean position. Thus, the motion of a helical spring is Simple Harmonic Motion.

Then, Time period of the spring $(T)$, \[T=2π \sqrt{\frac{y}{a}}\] and, in magnitude only, \[a=\frac{k}{m}y\] \[∴T=2π \sqrt{\frac{y}{\frac{k}{m}y}}\] \[T=2π \sqrt{\frac{m}{k}}\]

This gives the time period of the spring.

**More on Simple Harmonic Motion**

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