Question

A spring is cut into three equal pieces and connected as shown in the figure. By what factor will the time period of oscillation change if a block is attached before and after?

Answer 1

Hint: We need to understand the relation between the spring constant of a material of the spring and the length of the spring to solve this problem. We can find the difference in time periods of the two situations easily by using the relation between them.

Complete step-by-step solution
The spring is a helical loop of a material, which can extend or compress on applying an external force. The amount of the extension is dependent on the stiffness of the given spring. The stiffer the spring, the lesser will be the extension of the spring under a given force. The stiffness of spring is given in terms of the stiffness constant or the spring constant ‘k’.
The spring constant is dependent on the length of the spring. The product of the spring constant and the length of the spring is always a constant for a given material. This means that the spring constant and the length of the spring are inversely proportional.
i.e.,
\[k\propto \dfrac{1}{l}\]
In the present situation, the time period of oscillation of a spring of length ‘l’ and spring constant ‘k’ is given as –
\[\begin{align}
  & T=\dfrac{2\pi }{\omega } \\
 & \therefore T=2\pi \sqrt{\dfrac{m}{k}} \\
\end{align}\]

Now, it is said that the spring is cut into three equal parts and arranged as given in the figure. We know from the relation between the spring constant and the length of the spring that the new spring constant will become –
\[\begin{align}
  & l'=\dfrac{l}{3} \\
 & \therefore k'=3k \\
\end{align}\]
Now, we can see from the figure that the springs 1 and 2 will together have 6k as spring constant and the total spring constant including the spring 3 will be –
\[\begin{align}
  & \dfrac{1}{{{k}_{new}}}=\dfrac{1}{k'}+\dfrac{1}{2k'} \\
 & \Rightarrow {{k}_{new}}=\dfrac{2k'}{3} \\
 & \therefore {{k}_{new}}=2k \\
\end{align}\]
So, we can find the new time period of the system of springs as –
\[\begin{align}
  & T'=\dfrac{2\pi }{\omega '} \\
 & \Rightarrow T'=2\pi \sqrt{\dfrac{m}{{{k}_{new}}}} \\
 & \Rightarrow T'=2\pi \sqrt{\dfrac{m}{2k}} \\
 & \therefore T'=\dfrac{1}{\sqrt{2}}T \\
\end{align}\]
This is the new time period of the system.
This is the required solution.

Note: The spring system is highly dependent on the length and the mass of the spring. The stiffness per unit length determines the spring constant value per unit length. So, a change in the length is inversely dependent on the stiffness constant or ‘k’.