Two bodies A and B of equal mass are suspended from two massless springs of spring constant k₁ and k₂, 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

To find the ratio of the maximum velocity of body A to body B, let's analyze the simple harmonic motion (SHM) of both systems.

Step 1: Expression for Maximum Velocity in SHM

For a body undergoing simple harmonic motion, the velocity at any displacement $x$ is given by:

$$v = \omega \sqrt{A^2 - x^2}$$

where:

$\omega$ is the angular frequency of oscillation,

$A$ is the amplitude of oscillation,

$x$ is the instantaneous displacement.

The maximum velocity occurs when $x = 0$ (i.e., at the mean position), given by:

$v_{max}=\omega A$

where $\omega$ is the angular frequency, which depends on the spring constant $k$ and mass $m$:

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

Thus, the maximum velocity is:

$v_{max}=A\sqrt{\frac{k}{m}}$

Step 2: Finding the Ratio of Maximum Velocities

For body A (with spring constant $k_1$):

$$v_{A,\max} = A \sqrt{\frac{k_1}{m}}$$

For body B (with spring constant $k_2$):

$$v_{B,\max} = A \sqrt{\frac{k_2}{m}}$$

Since the amplitudes are equal, the ratio of their maximum velocities is:

$$\frac{v_{A,\max}}{v_{B,\max}} = \frac{A \sqrt{k_1/m}}{A \sqrt{k_2/m}}$$ $$= \sqrt{\frac{k_1}{k_2}}$$

Final Answer:

$$\mathbf{\sqrt{\frac{k_1}{k_2}}}$$