JEE Physics 2025 Question with Solution

Given: Two bodies A and B have equal mass and equal amplitudes of oscillation. They are attached to springs of spring constants $k_1$ and $k_2$ respectively.

Find: The ratio of maximum velocity $\frac{v_A}{v_B}$.

For a spring-mass system in simple harmonic motion, the maximum velocity is

$$v_{\text{max}} = A\omega$$

where $A$ is the amplitude and $\omega$ is the angular frequency.

Also,

$$\omega = \sqrt{\frac{k}{m}}$$

Since the amplitudes of both bodies are equal,

$$\frac{v_A}{v_B} = \frac{A\omega_A}{A\omega_B} = \frac{\omega_A}{\omega_B}$$

Using $\omega = \sqrt{\frac{k}{m}}$ for each body,

$$\frac{v_A}{v_B} = \frac{\sqrt{\frac{k_1}{m}}}{\sqrt{\frac{k_2}{m}}} = \sqrt{\frac{k_1}{k_2}}$$

Therefore, the ratio of the maximum velocity of A to that of B is $\sqrt{\frac{k_1}{k_2}}$. Hence, the correct option is D.

Given: Equal masses and equal amplitudes for both oscillators.

Find: The ratio of maximum speeds.

Since

$$v_{\text{max}} = A\omega$$

and the amplitudes are equal, maximum speed is directly proportional to angular frequency.

For a spring-mass system,

$$\omega \propto \sqrt{k}$$

when mass is the same.

Therefore,

$$\frac{v_A}{v_B} = \sqrt{\frac{k_1}{k_2}}$$

So, the correct option is D.