Question

When a mass \( m \) is connected individually to the springs \( s_1 \) and \( s_2 \), the oscillation frequencies are \( v_1 \) and \( v_2 \). If the same mass is attached to the two springs as shown in the figure, the oscillation frequency would be:

• A. \( v_1 + v_2 \)
• B. \( \sqrt{v_1^2 + v_2^2} \)
• C. \( \left( \frac{1}{v_1} + \frac{1}{v_2} \right)^{-1} \)
• D. \( \sqrt{v_1^2 - v_2^2} \)

Hint:

For two springs in parallel, the effective spring constant is \( k_{\text{eff}} = k_1 + k_2 \), and the resultant frequency follows \( v = \sqrt{v_1^2 + v_2^2} \).

Answer 1

The Correct Option is B

Step 1: Understanding the Given System - When a mass \( m \) is connected to a single spring of spring constant \( k \), the oscillation frequency is given by: \[ v = \frac{1}{2\pi} \sqrt{\frac{k}{m}}. \] - Here, the two springs are connected in parallel, meaning they act together to provide an effective restoring force.
Step 2: Finding the Effective Spring Constant For two parallel springs with constants \( k_1 \) and \( k_2 \), the equivalent spring constant is: \[ k_{\text{eff}} = k_1 + k_2. \] Using the frequency formula: \[ v_1 = \frac{1}{2\pi} \sqrt{\frac{k_1}{m}}, \quad v_2 = \frac{1}{2\pi} \sqrt{\frac{k_2}{m}}. \] Squaring both equations: \[ k_1 = 4\pi^2 m v_1^2, \quad k_2 = 4\pi^2 m v_2^2. \] The effective frequency for the parallel combination is: \[ v_{\text{eff}} = \frac{1}{2\pi} \sqrt{\frac{k_1 + k_2}{m}}. \] Substituting \( k_1 + k_2 \): \[ v_{\text{eff}} = \frac{1}{2\pi} \sqrt{\frac{4\pi^2 m v_1^2 + 4\pi^2 m v_2^2}{m}}. \] \[ v_{\text{eff}} = \sqrt{v_1^2 + v_2^2}. \] Thus, the correct answer is: \[ \boxed{\sqrt{v_1^2 + v_2^2}}. \]