Question

The potential energy of a simple harmonic oscillator, when the particle is half way to its end point is :
A) $\dfrac{1}{4}E$
B) $\dfrac{1}{2}E$
C) $\dfrac{2}{3}E$
D) $\dfrac{1}{8}E$

Answer 1

Hint:
As it is asked to calculate the potential energy from its end to the halfway point, you can solve this question by using the relationship between potential energy and the displacement of the particle.


Formula used:
The potential energy
$U = \dfrac{1}{2}m{\omega ^2}{x^2}$


Complete step by step solution:

The potential energy is the power that an object can store due to its position in relation to other things, internal tensions, electric charge, or other circumstances.
$U = \dfrac{1}{2}m{\omega ^2}{x^2}$

Half of the particle's amplitude, or when the particle is halfway to its end point, indicates this:
$x = \dfrac{a}{2}$
Now, substitute the value of displacement from mean position in the potential energy, then we obtain:
$U = \dfrac{1}{2}m{\omega ^2}{\left( {\dfrac{a}{2}} \right)^2} \\$
 $\Rightarrow U = \dfrac{1}{4}\left( {\dfrac{1}{2}m{\omega ^2}{a^2}} \right) \\$
 $\Rightarrow U = \dfrac{1}{4}E \\$
Thus, the correct option is:(A) $\dfrac{1}{4}E$






Note:
It should be noted that as a particle is moving from its extreme position to its mean position, the potential energy is going to decrease.