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 the amplitude of A and B is:

The correct option is C

$\sqrt{\frac{K_2}{K_1}}$

Step-1: Given data:

The maximum velocity of the particle executing SHM is given by, $V_{max}=A\omega$ …..(1)

In equation (1) A is the amplitude and $\omega =\sqrt{\frac{k}{m}}$

Step-2: Calculation:

Let us take the amplitude of particles A and B be $A_1$ and $A_2$ respectively.

Since maximum velocities are equal, we can write

$A_1{\omega }_1=A_2{\omega }_2$ $\left[\omega =\sqrt{\frac{k}{m}}\right]$

$A_1\times \sqrt{\frac{k_1}{m_1}}=A_2\times \sqrt{\frac{k_2}{m_2}}$

$\therefore m_1=m_2\Rightarrow \frac{A_1}{A_2}=\sqrt{\frac{k_2}{k_1}}$

Ratio of the amplitude of A and B = $\sqrt{\frac{k_2}{k_1}}$

Hence, Option (C) is correct.