Question

If the potential energy at a displacement \(x\) is numerically equal to the square root of potential energy at a displacement \(y\), then (where \(\omega\) is angular velocity):

• A. \(\frac{x^2}{y} = \omega \sqrt{\frac{m}{2}}\)
• B. \(\frac{y}{x^2} = \omega \sqrt{\frac{m}{2}}\)
• C. \(\frac{y}{x^2} = \frac{1}{\omega}\sqrt{\frac{m}{2}}\)
• D. \(\frac{x^2}{y} = \frac{1}{\omega}\sqrt{\frac{1}{2m}}\)

Hint:

In SHM, always use \(U = \frac{1}{2}m\omega^2 x^2\) for potential energy based relations.

Answer 1

The Correct Option is B

Step 1: Write expression of potential energy in SHM.
For SHM: \[ U = \frac{1}{2}m\omega^2 x^2 \]

Step 2: Apply given condition.
Given: \[ U_x = \sqrt{U_y} \] So, \[ \frac{1}{2}m\omega^2 x^2 = \sqrt{\frac{1}{2}m\omega^2 y^2} \]

Step 3: Simplify RHS.
\[ \sqrt{\frac{1}{2}m\omega^2 y^2} = y \sqrt{\frac{1}{2}m\omega^2} \]

Step 4: Equate both sides.
\[ \frac{1}{2}m\omega^2 x^2 = y \sqrt{\frac{1}{2}m\omega^2} \]

Step 5: Rearrange terms.
Divide both sides by \(x^2\): \[ \frac{y}{x^2} = \frac{\frac{1}{2}m\omega^2}{\sqrt{\frac{1}{2}m\omega^2}} \]

Step 6: Final simplification.
\[ \frac{y}{x^2} = \omega \sqrt{\frac{m}{2}} \]
Final Answer: \[ \boxed{\frac{y}{x^2} = \omega \sqrt{\frac{m}{2}}} \]