Question

Two equal masses hung from two massless springs of spring constants $k_1$ and $k_2$ have equal maximum velocity when executing SHM. The ratio of their amplitudes is

• A. $\sqrt{\frac{k_1}{k_2}}$
• B. $\frac{k_1}{k_2}$
• C. $\frac{k_2}{k_1}$
• D. $\sqrt{\frac{k_2}{k_1}}$
• E. $\frac{k_1^2}{k_2^2}$

Hint:

Higher spring constant → smaller amplitude for same maximum velocity.

Answer 1

The Correct Option is D

Concept: Maximum velocity in SHM: \[ v_{\max} = A \omega \] \[ \omega = \sqrt{\frac{k}{m}} \]

Step 1: Write for both: \[ v_1 = A_1 \sqrt{\frac{k_1}{m}}, \quad v_2 = A_2 \sqrt{\frac{k_2}{m}} \]

Step 2: Given $v_1 = v_2$: \[ A_1 \sqrt{k_1} = A_2 \sqrt{k_2} \]

Step 3: Rearranging: \[ \frac{A_1}{A_2} = \sqrt{\frac{k_2}{k_1}} \] Final Answer: \[ \sqrt{\frac{k_2}{k_1}} \]