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: Understanding the Concept:
In SHM the maximum velocity occurs at the mean position and equals amplitude times angular frequency. For a spring-mass system, \(\omega = \sqrt{K/m}\).

Step 2: Key Formula or Approach:
\[ v_{max} = A\omega = A\sqrt{\frac Km} \]

Step 3: Detailed Explanation:
The masses of A and B are equal, so write for each particle:
\[ A_A\sqrt{\frac{K_A}{m}} = A_B\sqrt{\frac{K_B}{m}} \]
Cancel \(m\):
\[ A_A\sqrt{K_A} = A_B\sqrt{K_B} \]
\[ \frac{A_B}{A_A} = \frac{\sqrt{K_A}}{\sqrt{K_B}} \]
So the ratio is \(\sqrt{K_A}:\sqrt{K_B}\). The stiffer spring gives the smaller amplitude. Option (D) is the inverse of this ratio, and options (A) and (C) forget the square root.

Final Answer:
The ratio of the amplitude of B to that of A is \(\sqrt{K_A}:\sqrt{K_B}\), option (B). \[ \boxed{\sqrt{K_A}:\sqrt{K_B} \text{ (B)}} \]