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: Potential energy in SHM.
U = ½ m ω² x².

Step 2: Applying the given condition.
U_x = √(U_y) → ½ m ω² x² = √(½ m ω² y²).

Step 3: Simplifying the square root.
√(½ m ω² y²) = y √(½ m ω²).

Step 4: Equating both sides.
½ m ω² x² = y √(½ m ω²).

Step 5: Isolating y/x².
y/x² = (½ m ω²) / √(½ m ω²) = ω √(m/2).

Step 6: Conclusion.
The relation is y/x² = ω √(m/2).