JEE MainPhysicsOscillations
A particle executes simple harmonic motion (SHM) and is at its extreme positive position at time t = 0 . Which of the following descriptions best represents the graph of the difference between its total mechanical energy and its potential energy, plotted as a function of time t ?
Options
- AA curve starting from a maximum positive value at t=0 , remaining non-negative, and oscillating periodically
- BA curve starting from zero at t=0 , remaining non-negative, and oscillating periodically to a maximum positive
- CAn inverted parabola opening downwards, symmetric about the vertical axis at t=0
- DA sinusoidal curve starting from zero at t=0 and oscillating symmetrically between positive and negative value
Correct answer
B. A curve starting from zero at t=0 , remaining non-negative, and oscillating periodically to a maximum positive
Step-by-step solution
The difference between the total mechanical energy ( E ) and the potential energy ( U ) of a particle in SHM is its kinetic energy ( K ). K = E - U The particle starts from the extreme positive position at t = 0 . Therefore, its position as a function of time is given by: x(t) = A ( t) The velocity of the particle is: v(t) = dx dt = -A ( t) The kinetic energy as a function of time is: K(t) = 1 2 mv^2 = 1 2 mA^2 ^2 ^2( t) At t = 0 , K(0) = 0 . Since kinetic energy depends on the square of the velocity, it is always