Question

Two particles A and B of equal masses are suspended from two massless springs of spring constants \(K_A\) and \(K_B\) respectively. The maximum velocities of the particles during oscillations are equal. The ratio of the amplitude of 'B' to that of 'A' is

• A. \(K_A:K_B\)
• B. \(\sqrt{K_A}:\sqrt{K_B}\)
• C. \(K_B:K_A\)
• D. \(\sqrt{K_B}:\sqrt{K_A}\)

Hint:

v_max = A omega = A sqrt(K/m). Equal masses and equal v_max give A proportional to 1/sqrt(K).

Answer 1

The Correct Option is B

Step 1: Use energy:
Maximum kinetic energy equals maximum potential energy: $\frac12mv_{max}^2 = \frac12KA^2$. So $v_{max} = A\sqrt{K/m}$.

Step 2: Equal masses and equal speeds:
Equal $m$ and equal $v_{max}$ give equal $KA^2$ for the two particles: $K_AA_A^2 = K_BA_B^2$.

Step 3: Solve for the ratio:
$\frac{A_B^2}{A_A^2} = \frac{K_A}{K_B}$, so $\frac{A_B}{A_A} = \sqrt{\frac{K_A}{K_B}}$.

Final Answer:
Option (B). \[ \boxed{\sqrt{K_A}:\sqrt{K_B} \text{ (B)}} \]