When the potential energy of a particle executing simple harmonic motion is one fourth of its maximum value during the oscillation the displacement of the particle from the equilibrium position in terms of its amplitude $a$ is
A) $\dfrac{a}{2}$
B) $\dfrac{a}{4}$
C) $\dfrac{{2a}}{3}$
D) None of these

Hint:
Here in this question, it is given that the potential energy of the particle is one fourth of the maximum value of the total potential energy. From which we have to find the value of $x$ as in the terms of which the term of its amplitude will be $a$ .


Formula used :
We are aware that the potential energy formula is,
$P.E. = \dfrac{1}{2}k{x^2}$


Complete step by step solution:

As for the value of Potential energy for the maximum,
$P.E{._{\max }} = \dfrac{1}{2}k{a^2}$
As given in the question that the maximum value is one fourth to the Potential energy of a particle so executing the values as said in the question,
$\dfrac{1}{2}k{x^2} = \dfrac{1}{4} \times \dfrac{1}{2}k{A^2}$
As we need the value of $x$ , we rearrange the values as for the value of $x$ ,
$x = \dfrac{a}{2}$
As a result, the value of $x$ as from the full conversion with Potential energy is $x = \dfrac{a}{2}$ .
Therefore, from the whole conclusion the correct answer is $\dfrac{a}{2}$ .
Hence, the correct option is (A).



Hence the correct answer is Option(A).













Note:
Here a term used for simple harmonic motion we are going to know what is simple harmonic motion. Simple harmonic motion is described as a periodic motion of a point in a straight line whose acceleration always points in the direction of a fixed point on the line and whose distance from that point is proportional.