A particle executes simple harmonic motion along the straight line with an amplitude A. The potential energy is maximum when the displacement is

(A) $ \pm A$
(B) Zero
(C) $ \pm \dfrac{A}{2}$
(D) $ \pm \dfrac{A}{{\sqrt 2 }}$

Hint:
First start with finding the relation of the potential energy of a particle executing simple harmonic motion (S.H.M.) and try to find out which of the given options is fit in that relation and finally get the right answer and you can use the method of elimination and can eliminate the wrong option one by one.






Formula used :
Potential energy : $P.E. = 1/2{\text{ }}k{x^2}\; = {\text{ }}1/2\;m\;{\omega ^2}{x^2}$


Complete step by step solution:
Potential energy is the energy possessed by the particle when the particle is at rest.
Now when the particle is executing simple harmonic motion at a distance x from the mean position.
The force acting will be $F = - kx$
Now the work done will be:
$dW = - fdx$
After solving, we get;
Total work done, $W = 1/2{\text{ }}K{\text{ }}{x^2}\; = {\text{ }}1/2\;m\;{\omega ^2}{x^2}$
The total work done here will get stored in the form of potential energy.
So, Potential energy : $P.E. = 1/2{\text{ }}k{x^2}\; = {\text{ }}1/2\;m\;{\omega ^2}{x^2}$
Now, P.E. will be maximum when the value of x will be maximum: x will be maximum at $x = \pm A$

Hence the correct answer is Option(A).










Note:
Find the total energy in case of a particle moving in a simple harmonic motion which is equal to the sum of kinetic energy and potential energy of the particle. Put all the values in the formula of the kinetic and potential energy carefully and get the required answer.