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: Energy form
Total energy is constant: $\frac{1}{2}mV_1^2 + \frac{1}{2}m\omega^2x_1^2 = \frac{1}{2}mV_2^2 + \frac{1}{2}m\omega^2x_2^2$.

Step 2: Rearrange
$V_2^2 - V_1^2 = \omega^2(x_1^2 - x_2^2)$.

Step 3: Solve
$\omega^2 = \dfrac{V_2^2 - V_1^2}{x_1^2 - x_2^2}$.

Step 4: Sense check
A particle moving closer to the mean position ($x_2 < x_1$) is faster ($V_2 > V_1$), so both numerator and denominator are positive, as required.

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