Question

The potential energy of a simple harmonic oscillator when the particle is half way to its end point is (where $ E $ is the total energy):

• A. $ \frac{2}{3} E $
• B. $ \frac{1}{8} E $
• C. $ \frac{1}{4} E $
• D. $ \frac{1}{2} E $

Hint:

Remember that Potential Energy in SHM is proportional to the square of displacement ($ U \propto x^2 $). If the distance is halved, the energy becomes $ (1/2)^2 = 1/4 $ of the maximum energy.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
In Simple Harmonic Motion (SHM), the total energy \( E \) is the sum of kinetic and potential energy. At the "end points" (amplitude \( A \)), the particle stops momentarily, meaning kinetic energy is zero and all energy is potential.
Key Formula or Approach:
Potential Energy \( U = \frac{1}{2}kx^2 \).
Total Energy \( E = \frac{1}{2}kA^2 \).
Step 2: Detailed Explanation:
The problem states the particle is "half way to its end point". This means the displacement \( x = A/2 \).
Substitute \( x \) into the potential energy formula:
\[ U = \frac{1}{2}k \left(\frac{A}{2}\right)^2 = \frac{1}{2}k \frac{A^2}{4} \]
We can rewrite this as:
\[ U = \frac{1}{4} \left( \frac{1}{2}kA^2 \right) \]
Since \( \frac{1}{2}kA^2 = E \), we get:
\[ U = \frac{1}{4}E \]
Step 3: Final Answer:
The potential energy is \( \frac{1}{4}E \).