Question

A plot of potential energy v/s kinetic energy of a particle executing SHM gives us a straight line:
A) passing through the origin
B) with positive slope
C) with intercepts on both the axes
D) parallel to either of the axis

Answer 1

The potential energy of a particle executing SHM will be due to its tendency to move towards its mean position. The kinetic energy of the particle will be due to its finite velocity about its mean position.

Complete step by step answer:
In a simple harmonic oscillator (SHM), the total energy of the oscillator will remain constant. Since the total energy of the oscillator will be the sum of kinetic and potential energy of the oscillator.
So, we can write that
$K + U = c$ where $c$ is a constant which is the total energy of the system.
This equation is similar to the equation of a line
$\dfrac{x}{a} + \dfrac{y}{b} = 1$ where $a$ and $b$ are the x and y intercepts.
Hence the equation $K + U = c$ , is a line with intercepts at both the axes. This implies that both the kinetic and the potential energy of the oscillator will be zero at some point during its oscillation. The kinetic energy will be maximum when the oscillator is at its mean position where the oscillator will have maximum velocity. The potential energy of the oscillator will be maximum when the oscillator is at the extreme points away from its mean position i.e., its amplitude.
So, the correct choice is option (C).

Note:
The kinetic and the potential energy of an oscillator vary at a frequency which is twice the frequency of the oscillator. The total energy of the oscillator will remain constant during its entire oscillatory motion which implies that potential and kinetic energy interchange with each other. When the potential energy is maximum, kinetic energy is minimum and vice versa.