Question

The variation of displacement with time of a particle executing free simple harmonic motion is shown in the figure. 
The potential energy U(x) versus time (t) plot of the particle is correctly shown in figure : 

• A. A
• B. B
• C. C
• D. D

Hint:

For SHM, remember the energy transformations. Potential energy is max at extremes, zero at the mean. Kinetic energy is max at the mean, zero at extremes. Total energy is constant. The frequency of both potential and kinetic energy oscillations is twice the frequency of the displacement oscillation.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given a displacement-time (\(x-t\)) graph for a particle in Simple Harmonic Motion (SHM). We need to identify the correct potential energy-time (\(U-t\)) graph for the same motion.
Step 2: Key Formula or Approach:
The potential energy (\(U\)) of a particle in SHM is given by \(U = \frac{1}{2}kx^2\), where \(k\) is the spring constant and \(x\) is the displacement from the mean position.
The displacement \(x\) as a function of time \(t\) for the given graph is \(x(t) = A\sin(\omega t)\).
Substituting \(x(t)\) into the potential energy formula gives \(U(t)\).
Step 3: Detailed Explanation:
From the formula \(U = \frac{1}{2}kx^2\), we can deduce the key features of the potential energy graph: 1. Non-negativity: Since \(k\) is positive and \(x^2\) is always non-negative, the potential energy \(U\) must always be greater than or equal to zero (\(U \ge 0\)). This eliminates any graph that goes below the time axis (like option A).
2. Value at mean position: At the mean position, the displacement is zero (\(x=0\)). The given \(x-t\) graph shows this at points O, B, etc. At these points, the potential energy must be zero: \(U = \frac{1}{2}k(0)^2 = 0\). This eliminates graphs where the minimum value is greater than zero (like option C).
3. Value at extreme positions: At the extreme positions (amplitude), the displacement is maximum (\(x = \pm A\)). The given \(x-t\) graph shows this at points A (maximum positive displacement) and C (maximum negative displacement). At these points, the potential energy is maximum: \(U_{max} = \frac{1}{2}k(\pm A)^2 = \frac{1}{2}kA^2\).
4. Frequency: The displacement is \(x(t) = A\sin(\omega t)\). The potential energy is \(U(t) = \frac{1}{2}k(A\sin(\omega t))^2 = \frac{1}{2}kA^2\sin^2(\omega t)\). Using the identity \(\sin^2\theta = \frac{1-\cos(2\theta)}{2}\), we get \(U(t) = \frac{1}{4}kA^2(1 - \cos(2\omega t))\). The angular frequency of the potential energy oscillation is \(2\omega\), which is double the frequency of the displacement oscillation. This means that for every one full cycle of displacement, the potential energy completes two full cycles.
Step 4: Final Answer:
Let's check the options against these criteria: - Option (A) is incorrect because energy cannot be negative. - Option (C) is incorrect because potential energy is zero at the mean position. - Option (B) correctly shows that \(U \ge 0\), \(U=0\) at points corresponding to O and B on the \(x-t\) graph, and \(U=U_{max}\) at points corresponding to A and C. It also shows two energy cycles for one displacement cycle. - Option (D) might have similar features, but Graph (B) is the standard representation matching all criteria.
Therefore, the graph with ID 86435168204 is the correct representation.