Question

Two bodies A and B of equal mass are suspended from two massless springs of spring constant \( k_1 \) and \( k_2 \), respectively. If the bodies oscillate vertically such that their amplitudes are equal, the ratio of the maximum velocity of A to the maximum velocity of B is:

• A. \( \frac{k_1}{k_2} \)
• B. \( \frac{k_2}{k_1} \)
• C. \( \sqrt{\frac{k_2}{k_1}} \)
• D. \( \sqrt{\frac{k_1}{k_2}} \)

Hint:

In simple harmonic motion, the maximum velocity of an oscillating body is directly proportional to the amplitude and angular frequency. For systems with equal amplitudes, the ratio of maximum velocities depends on the square root of the ratio of spring constants.

Answer 1

The Correct Option is D

For a spring-mass system undergoing simple harmonic motion, the maximum velocity is given by: \[ v_{\text{max}} = A \omega \] where \( A \) is the amplitude, and \( \omega \) is the angular frequency. The angular frequency \( \omega \) is related to the spring constant and mass by: \[ \omega = \sqrt{\frac{k}{m}} \] Given that the amplitudes of both bodies are equal, we can write the ratio of the maximum velocities for bodies A and B as: \[ \frac{v_A}{v_B} = \frac{A \omega_A}{A \omega_B} = \frac{\omega_A}{\omega_B} = \sqrt{\frac{k_1}{k_2}} \] Thus, the ratio of the maximum velocity of A to that of B is \( \sqrt{\frac{k_1}{k_2}} \).