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

The maximum velocity \( v_{\text{max}} \) of a spring-mass system in simple harmonic motion is \( v_{\text{max}} = A \omega \), where \( A \) is the amplitude and \( \omega \) is the angular frequency. The angular frequency \( \omega \) is defined as \( \omega = \sqrt{\frac{k}{m}} \), with \( k \) being the spring constant and \( m \) the mass. For two bodies with identical amplitudes, the ratio of their maximum velocities is \( \frac{v_A}{v_B} = \frac{A \omega_A}{A \omega_B} = \frac{\omega_A}{\omega_B} = \sqrt{\frac{k_1}{k_2}} \). Consequently, the ratio of the maximum velocity of body A to that of body B equals \( \sqrt{\frac{k_1}{k_2}} \).