Question

Two bodies M and N of equal mass are suspended from two separate massless spring of force constant $k_1$ and $k_2$ respectively. If the two bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitude of vibration of M to that of N is?

• A. $\frac{k_1}{k_2}$
• B. $\sqrt{\frac{k_1}{k_2}}$
• C. $\frac{k_2}{k_1}$
• D. $\sqrt{\frac{k_2}{k_1}}$

Answer 1

The correct option is D $\sqrt{\frac{k_2}{k_1}}$

$V_{max}=A\omega$

$⟹{V_{max}}_1={V_{max}}_2$

$⟹A_1{\omega }_1=A_2{\omega }_2$

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

Now, $K_1=m{\omega }_1^2$

$⟹{\omega }_1=\sqrt{\frac{K_1}{m}}$

and $K_2=m{\omega }_2^2$

$⟹{\omega }_2=\sqrt{\frac{K_2}{m}}$

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