Question

A massless spring of length \( l \) and spring constant \( k \) oscillates with a time period \( T \) when loaded with a mass \( m \). The spring is now cut into three equal parts and connected in parallel. The frequency of oscillation of the combination when it is loaded with a mass \( 4m \) is: \vspace{0.5cm}

• A. \( \frac{2}{T} \)
• B. \( \frac{2}{3T} \)
• C. \( \frac{3}{T} \)
• D. \( \frac{3}{2T} \) \vspace{0.5cm}

Hint:

For a spring cut into \( n \) equal parts, the new spring constant is \( k' = nk \), and when connected in parallel, the effective spring constant is \( k_{\text{eff}} = n^2 k \).

Answer 1

The Correct Option is D

Step 1: Understanding the Effect of Cutting the Spring The time period for a mass-spring system is given by: \[ T = 2\pi \sqrt{\frac{m}{k}} \] When the spring is cut into three equal parts, each segment will have a new spring constant: \[ k' = 3k \] Since they are connected in parallel, the effective spring constant becomes: \[ k_{\text{eff}} = 3k + 3k + 3k = 9k \] \vspace{0.5cm} Step 2: Finding the New Time Period The new time period when a mass \( 4m \) is attached: \[ T' = 2\pi \sqrt{\frac{4m}{9k}} \] \[ T' = 2\pi \times \frac{2}{3} \sqrt{\frac{m}{k}} \] \[ T' = \frac{2}{3} T \] \vspace{0.5cm} Step 3: Finding the Frequency Frequency is the reciprocal of the time period: \[ f' = \frac{1}{T'} = \frac{1}{\frac{2}{3} T} = \frac{3}{2T} \] Thus, the frequency of oscillation of the combination is: \[ \mathbf{\frac{3}{2T}} \] \vspace{0.5cm}