Physics - SHM Question with Solution | TestHub

Let the masses of the particles be $m$.

The angular frequencies of oscillations for particles A and B are ${\omega }_A=\sqrt{\frac{K_A}{m}}$ and ${\omega }_B=\sqrt{\frac{K_B}{m}}$.

The maximum velocity of a particle in simple harmonic motion is given by $v_{max}=A\omega$, where $A$ is the amplitude.

Given that the maximum velocities are equal:

$A_A{\omega }_A=A_B{\omega }_B$

$A_A\sqrt{\frac{K_A}{m}}=A_B\sqrt{\frac{K_B}{m}}$

$A_A\sqrt{K_A}=A_B\sqrt{K_B}$

$\frac{A_B}{A_A}=\frac{\sqrt{K_A}}{\sqrt{K_B}}$

The ratio of the amplitude of B to that of A is $\sqrt{K_A}:\sqrt{K_B}$.

Answer: $\sqrt{K_A}:\sqrt{K_B}$