Question

A particle executing linear S.H.M. has velocities \(V_1\) and \(V_2\) at distance \(x_1\) and \(x_2\) respectively, from the mean position, its angular velocity is

• A. \(\sqrt{\frac{V_2^2-V_1^2}{x_1^2-x_2^2}}\)
• B. \(\sqrt{\frac{V_2^2-V_1^2}{x_2^2-x_1^2}}\)
• C. \(\sqrt{\frac{V_2^2-V_1^2}{x_1x_2}}\)
• D. \(\sqrt{\frac{V_1V_2}{x_1+x_2}}\)

Hint:

Use v squared = omega squared (A squared - x squared) at the two positions and subtract.

Answer 1

The Correct Option is A

Step 1: Understand the concept
For linear SHM, the speed at displacement \(x\) is \(v^2 = \omega^2(A^2 - x^2)\).

Step 2: Write for the two positions
\(V_1^2 = \omega^2(A^2 - x_1^2)\) and \(V_2^2 = \omega^2(A^2 - x_2^2)\).

Step 3: Subtract
\[ V_2^2 - V_1^2 = \omega^2(x_1^2 - x_2^2) \]

Step 4: Solve
\[ \omega = \sqrt{\frac{V_2^2 - V_1^2}{x_1^2 - x_2^2}} \]
Option (A). Option (B) has the denominator in the opposite order and would give a negative value under the root whenever \(V_2 > V_1\) and \(x_1 > x_2\).

Final Answer:
Angular velocity is sqrt((V2^2 - V1^2)/(x1^2 - x2^2)). This is option (A). \[ \boxed{\text{(A) }\sqrt{\frac{V_2^2-V_1^2}{x_1^2-x_2^2}}} \]