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. \[ \text{Graph 1: } \]
  
• B. \[ \text{Graph 2: } \] 
  
• C. \[ \text{Graph 3: } \] 
  
• D. \[ \text{Graph 4: } \] 
  

Hint:

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.

Answer 1

The Correct Option is A

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.