Physics - SHM Question with Solution | TestHub

Let the mass of both bodies be $m$.

The angular frequencies of oscillation for bodies A and B are ${\omega }_1=\sqrt{\frac{K_1}{m}}$ and ${\omega }_2=\sqrt{\frac{K_2}{m}}$ respectively.

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

Given that the maximum velocities of both bodies are equal:

$A_1{\omega }_1=A_2{\omega }_2$

Substituting the expressions for ${\omega }_1$ and ${\omega }_2$:

$A_1\sqrt{\frac{K_1}{m}}=A_2\sqrt{\frac{K_2}{m}}$

$A_1\sqrt{K_1}=A_2\sqrt{K_2}$

The ratio of the amplitude of B to that of A is:

$\frac{A_2}{A_1}=\sqrt{\frac{K_1}{K_2}}$

Answer: $\sqrt{\frac{{\mathrm{K}}_1}{{\mathrm{K}}_2}}$