Question

Five identical springs are used in the three configurations as shown in figure. The time periods of vertical oscillations in configurations (a), (b) and (c) are in the ratio:
\includegraphics[width=0.5\linewidth]{PH39.png}

• A. \( 1:\sqrt{2}:\frac{1}{\sqrt{2}} \)
• B. \( 2:\sqrt{2}:\frac{1}{\sqrt{2}} \)
• C. \( \frac{1}{\sqrt{2}}:2:1 \)
• D. \( 2:\frac{1}{\sqrt{2}}:1 \)

Hint:

For springs in series, the effective spring constant is given by \( \frac{1}{k_{eff}} = \sum \frac{1}{k_i} \). For springs in parallel, the effective spring constant is given by \( k_{eff} = \sum k_i \). The time period of a spring-mass system is \( T = 2\pi \sqrt{\frac{m}{k}} \).

Answer 1

The Correct Option is A

Step 1: The time period of a spring-mass system is given by the formula: \[ T = 2\pi \sqrt{\frac{m}{k}} \] where: \begin{itemize} \item $T$ is the time period \item $m$ is the mass \item $k$ is the spring constant \end{itemize} For the given systems: \begin{enumerate} \item System (a): \[ T_a = 2\pi \sqrt{\frac{m}{k}} \] \item System (b): The spring constant is halved, i.e., $k' = \frac{k}{2}$. Therefore, \[ T_b = 2\pi \sqrt{\frac{m}{k/2}} = 2\pi \sqrt{\frac{2m}{k}} = \sqrt{2} \left( 2\pi \sqrt{\frac{m}{k}} \right) = \sqrt{2} T_a \] \item System (c): The spring constant is doubled, i.e., $k'' = 2k$. Therefore, \[ T_c = 2\pi \sqrt{\frac{m}{2k}} = \frac{1}{\sqrt{2}} \left( 2\pi \sqrt{\frac{m}{k}} \right) = \frac{1}{\sqrt{2}} T_a \] \end{enumerate} Thus, the ratio of the time periods is: \[ T_a : T_b : T_c = 1 : \sqrt{2} : \frac{1}{\sqrt{2}} \]