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. 

 

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

Hint:

Series makes the system "softer" (smaller \(k\), larger \(T\)), while parallel makes it "stiffer" (larger \(k\), smaller \(T\)).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept: 
The time period of a mass-spring system is determined by the equivalent spring constant $k_{\text{eq}}$ of the given configuration. Series and parallel combinations of springs alter the equivalent stiffness. 
Step 2: Key Formula or Approach: 
The time period of oscillation is given by: \[ T = 2\pi \sqrt{\frac{m}{k_{\text{eq}}}} \] For springs in series: $\frac{1}{k_{\text{eq}}} = \frac{1}{k_1} + \frac{1}{k_2}$ 
For springs in parallel: $k_{\text{eq}} = k_1 + k_2$ 
Step 3: Detailed Explanation: 
Let the spring constant of each identical spring be $k$. 
Configuration (a): A single spring is attached to the mass. \[ k_a = k \] \[ T_a = 2\pi \sqrt{\frac{m}{k}} \] Configuration (b): Two springs are connected in series. \[ \frac{1}{k_b} = \frac{1}{k} + \frac{1}{k} = \frac{2}{k} \implies k_b = \frac{k}{2} \] \[ T_b = 2\pi \sqrt{\frac{m}{k/2}} = \sqrt{2} \left( 2\pi \sqrt{\frac{m}{k}} \right) = \sqrt{2} T_a \] Configuration (c): Two springs are connected in parallel. \[ k_c = k + k = 2k \] \[ T_c = 2\pi \sqrt{\frac{m}{2k}} = \frac{1}{\sqrt{2}} \left( 2\pi \sqrt{\frac{m}{k}} \right) = \frac{1}{\sqrt{2}} T_a \] Taking the ratio of their time periods: \[ T_a : T_b : T_c = T_a : \sqrt{2} T_a : \frac{1}{\sqrt{2}} T_a = 1 : \sqrt{2} : \frac{1}{\sqrt{2}} \] Step 4: Final Answer: 
The ratio of the time periods is $1 : \sqrt{2} : \frac{1}{\sqrt{2}}$.