Question

A particle performs S.H.M. Its potential energies are \( U_1 \) and \( U_2 \) at displacements \( x_1 \) and \( x_2 \) respectively. At displacement \( x_1 + x_2 \), its potential energy \( U \) is

• A. \( \sqrt{U_1} + \sqrt{U_2} \)
• B. \( U = (\sqrt{U_1} + \sqrt{U_2})^2 \)
• C. \( U = \sqrt{U_1} - \sqrt{U_2} \)
• D. \( U = (\sqrt{U_1} - \sqrt{U_2})^2 \)

Hint:

In S.H.M., the potential energy depends on the square of the displacement. When combining displacements, apply the correct formula for the potential energy at the new position.

Answer 1

The Correct Option is A

Step 1: Understanding potential energy in S.H.M.
In Simple Harmonic Motion (S.H.M.), the potential energy at a displacement \( x \) is given by: \[ U = \frac{1}{2} k x^2 \] Where \( k \) is the spring constant and \( x \) is the displacement from the equilibrium position. Step 2: Using the given displacements.
At displacement \( x_1 \), the potential energy is \( U_1 = \frac{1}{2} k x_1^2 \), and at displacement \( x_2 \), the potential energy is \( U_2 = \frac{1}{2} k x_2^2 \). For the displacement \( x_1 + x_2 \), the potential energy will be: \[ U = \frac{1}{2} k (x_1 + x_2)^2 \] Thus, the potential energy at the combined displacement is represented by the sum of the individual potential energies, hence the correct answer is (A) \( \sqrt{U_1} + \sqrt{U_2} \).