Question:
A particle oscillates along the \( x \)-axis according to the law, \( x(t) = x_0 \sin^2 \left( \frac{\pi t}{T} \right) \), where \( x_0 = 1 \, \text{m} \) and \( T \) is the time period of oscillation. The kinetic energy (\( K \)) of the particle as a function of \( x \) is correctly represented by the graph:
A particle oscillates along the \( x \)-axis according to the law, \( x(t) = x_0 \sin^2 \left( \frac{\pi t}{T} \right) \), where \( x_0 = 1 \, \text{m} \) and \( T \) is the time period of oscillation. The kinetic energy (\( K \)) of the particle as a function of \( x \) is correctly represented by the graph:
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To determine the kinetic energy as a function of displacement, start by differentiating the displacement function to find the velocity. Then square the velocity and use the formula for kinetic energy.
Updated On: Apr 19, 2026
\[ \text{Graph 1: } \]
\[ \text{Graph 2: } \]
\[ \text{Graph 3: } \]
\[ \text{Graph 4: } \]
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The Correct Option is A
Approach Solution - 1
The kinetic energy \( K \) of a particle is given by: \[ K = \frac{1}{2} m v^2, \] where \( v \) is the velocity of the particle. The velocity is the derivative of the displacement \( x(t) \) with respect to time: \[ v(t) = \frac{d}{dt} \left( x_0 \sin^2 \left( \frac{\pi t}{T} \right) \right) = 2x_0 \sin \left( \frac{\pi t}{T} \right) \cos \left( \frac{\pi t}{T} \right) \frac{\pi}{T}. \] Thus, the velocity is proportional to \( \sin \left( \frac{\pi t}{T} \right) \), and the kinetic energy is proportional to the square of the velocity, which results in a graph where the kinetic energy increases as the particle moves from the origin to its maximum displacement and decreases symmetrically thereafter.
Final Answer: Graph 1.
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Approach Solution -2
Step 1: Analyze the given motion equation.
The position of the particle as a function of time is given by:
\[ x(t) = x_0 \sin^2 \left( \frac{\pi t}{T} \right), \] where \( x_0 = 1 \, \text{m} \) and \( T \) is the time period of the oscillation. The function describes oscillatory motion along the \( x \)-axis.
Step 2: Derive the velocity of the particle.
To find the velocity \( v(t) \), we differentiate the position function \( x(t) \) with respect to time \( t \):
\[ v(t) = \frac{d}{dt} \left( x_0 \sin^2 \left( \frac{\pi t}{T} \right) \right). \] Using the chain rule:
\[ v(t) = 2x_0 \sin \left( \frac{\pi t}{T} \right) \cos \left( \frac{\pi t}{T} \right) \cdot \frac{\pi}{T}. \] Simplifying the expression:
\[ v(t) = \frac{\pi x_0}{T} \sin \left( \frac{\pi t}{T} \right) \cos \left( \frac{\pi t}{T} \right), \] and applying the identity \( \sin(2\theta) = 2\sin(\theta)\cos(\theta) \), we get:
\[ v(t) = \frac{\pi x_0}{2T} \sin \left( \frac{2\pi t}{T} \right). \]
Step 3: Calculate the kinetic energy.
The kinetic energy \( K \) of the particle is given by:
\[ K = \frac{1}{2} m v(t)^2. \] Substituting the expression for \( v(t) \):
\[ K = \frac{1}{2} m \left( \frac{\pi x_0}{2T} \sin \left( \frac{2\pi t}{T} \right) \right)^2. \] This shows that the kinetic energy depends on the square of the velocity, and hence the position of the particle.
Step 4: Analyze the graph of kinetic energy.
The graph of kinetic energy as a function of position \( x \) will follow a curve that starts from zero, increases as the particle moves towards the midpoint of the oscillation, and then decreases symmetrically as the particle reaches the extreme positions (where \( x = 0 \) again). This behavior is characteristic of a sinusoidal oscillation.
Step 5: Final answer.
The correct graph for the kinetic energy \( K \) as a function of position \( x \) is the one that shows a smooth curve, starting at zero, reaching a maximum at the midpoint of the oscillation, and then decreasing symmetrically towards zero again. The graph shown in the image correctly represents this behavior.
The position of the particle as a function of time is given by:
\[ x(t) = x_0 \sin^2 \left( \frac{\pi t}{T} \right), \] where \( x_0 = 1 \, \text{m} \) and \( T \) is the time period of the oscillation. The function describes oscillatory motion along the \( x \)-axis.
Step 2: Derive the velocity of the particle.
To find the velocity \( v(t) \), we differentiate the position function \( x(t) \) with respect to time \( t \):
\[ v(t) = \frac{d}{dt} \left( x_0 \sin^2 \left( \frac{\pi t}{T} \right) \right). \] Using the chain rule:
\[ v(t) = 2x_0 \sin \left( \frac{\pi t}{T} \right) \cos \left( \frac{\pi t}{T} \right) \cdot \frac{\pi}{T}. \] Simplifying the expression:
\[ v(t) = \frac{\pi x_0}{T} \sin \left( \frac{\pi t}{T} \right) \cos \left( \frac{\pi t}{T} \right), \] and applying the identity \( \sin(2\theta) = 2\sin(\theta)\cos(\theta) \), we get:
\[ v(t) = \frac{\pi x_0}{2T} \sin \left( \frac{2\pi t}{T} \right). \]
Step 3: Calculate the kinetic energy.
The kinetic energy \( K \) of the particle is given by:
\[ K = \frac{1}{2} m v(t)^2. \] Substituting the expression for \( v(t) \):
\[ K = \frac{1}{2} m \left( \frac{\pi x_0}{2T} \sin \left( \frac{2\pi t}{T} \right) \right)^2. \] This shows that the kinetic energy depends on the square of the velocity, and hence the position of the particle.
Step 4: Analyze the graph of kinetic energy.
The graph of kinetic energy as a function of position \( x \) will follow a curve that starts from zero, increases as the particle moves towards the midpoint of the oscillation, and then decreases symmetrically as the particle reaches the extreme positions (where \( x = 0 \) again). This behavior is characteristic of a sinusoidal oscillation.
Step 5: Final answer.
The correct graph for the kinetic energy \( K \) as a function of position \( x \) is the one that shows a smooth curve, starting at zero, reaching a maximum at the midpoint of the oscillation, and then decreasing symmetrically towards zero again. The graph shown in the image correctly represents this behavior.
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