Question

Two particles $A$ and $B$ of equal masses are suspended from two massless springs of spring constants $k_1$ and $k_2$ respectively. If the maximum velocities, during oscillations, are equal, the ratio of amplitudes of $A$ and $B$ is:

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

Answer 1

The Correct Option is C

To find the ratio of amplitudes of particles \(A\) and \(B\), given that their maximum velocities during oscillations are equal, let's start by analyzing the physics of the situation. Both masses have simple harmonic motion properties with spring constants \(k_1\) and \(k_2\).

1. The formula for the maximum velocity \(V_{\text{max}}\) in simple harmonic motion is given by: \(V_{\text{max}} = A \cdot \omega\) where \(A\) is the amplitude and \(\omega\) is the angular frequency.
2. Angular frequency \(\omega\) is given by: \(\omega = \sqrt{\frac{k}{m}}\), where \(k\) is the spring constant, and \(m\) is the mass.
3. For particle \(A\): \(V_{A\text{, max}} = A_1 \cdot \sqrt{\frac{k_1}{m}}\)
4. For particle \(B\): \(V_{B\text{, max}} = A_2 \cdot \sqrt{\frac{k_2}{m}}\)
5. Since the maximum velocities are equal: \(A_1 \cdot \sqrt{\frac{k_1}{m}} = A_2 \cdot \sqrt{\frac{k_2}{m}}\)
6. Simplifying this equality, we get: \(\frac{A_1}{A_2} = \sqrt{\frac{k_2}{k_1}}\)

This leads to the ratio of amplitudes of particles \(A\) and \(B\) being: \(\sqrt{\frac{k_2}{k_1}}\).

Thus, the correct answer is option (C).