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

Concept: In Simple Harmonic Motion (SHM), the total energy $ E $ of the oscillator is constant and is the sum of its kinetic energy and potential energy. The potential energy ($ U $) at a displacement $ x $ is given by: $$ U = \frac{1}{2} k x^2 $$ The total energy ($ E $) occurs when the displacement is at its maximum (amplitude $ A $), where kinetic energy is zero: $$ E = \frac{1}{2} k A^2 $$

Step 1: Defining the displacement.
The problem states the particle is "half way to its end point." The end point of an oscillator is its amplitude $ A $. Therefore, the displacement $ x $ is: $$ x = \frac{A}{2} $$

Step 2: Calculating Potential Energy at that point. Substitute $ x = \frac{A}{2} $ into the potential energy formula: $$ U = \frac{1}{2} k \left( \frac{A}{2} \right)^2 $$ $$ U = \frac{1}{2} k \left( \frac{A^2}{4} \right) $$ $$ U = \frac{1}{4} \left( \frac{1}{2} k A^2 \right) $$

Step 3: Relating to Total Energy. Since we know that $ E = \frac{1}{2} k A^2 $, we can substitute $ E $ into our expression for $ U $: $$ U = \frac{1}{4} E $$