Question

A particle in SHM has amplitude \( 2 \text{ m} \) and total energy \( 8 \text{ J} \). Its potential energy at displacement \( 1 \text{ m} \) is:

• A. \( 1 \text{ J} \)
• B. \( 2 \text{ J} \)
• C. \( 4 \text{ J} \)
• D. \( 8 \text{ J} \)

Hint:

For energy problems in SHM, remember that Potential Energy scales quadratically with displacement (\(U \propto x^2\)). If the displacement becomes half of the maximum amplitude (\(x = A/2\)), the potential energy will always drop to exactly one-fourth (\(1/4\)) of the total energy!

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
In SHM, the energy of the particle constantly shifts between potential and kinetic forms, but the total mechanical energy \(E\) remains constant.
Total energy is equal to the maximum potential energy (at maximum displacement/amplitude) or the maximum kinetic energy (at the mean position).
Both potential energy \(U\) and total energy \(E\) are proportional to the square of their respective displacements.
Key Formula or Approach:
Total Energy: \(E = \frac{1}{2} k A^2\)
Potential Energy at displacement \(x\): \(U = \frac{1}{2} k x^2\)
Relationship: \(U = E \cdot \left(\frac{x}{A}\right)^2\)
Step 2: Detailed Explanation:
Given:
Amplitude \(A = 2 \text{ m}\)
Total energy \(E = 8 \text{ J}\)
Displacement \(x = 1 \text{ m}\)
We can express the potential energy as a fraction of the total energy by taking the ratio of their formulas:
\[ \frac{U}{E} = \frac{\frac{1}{2}kx^2}{\frac{1}{2}kA^2} = \frac{x^2}{A^2} \]
Substituting the numerical values:
\[ U = E \cdot \frac{x^2}{A^2} = 8 \cdot \frac{1^2}{2^2} = 8 \cdot \frac{1}{4} = 2 \text{ J} \]
This means at half of the amplitude, the potential energy is exactly one-fourth of the total energy.
Step 3: Final Answer:
The potential energy at a displacement of 1 m is \(2 \text{ J}\).