Question

The potential energy of a particle executing simple harmonic motion (S.H.M.) is $2.5J$ , when its displacement is half of the amplitude. The total energy of the particle will be:

(A) $18J$
(B) $10J$
(C) $12J$
(D) $2.5J$

Answer 1

Hint:
First start with finding the relation of the potential energy and the total 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, 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, total energy is T.E. = Kinetic energy + potential energy
$T.E = 1/2{\text{ }}K{\text{ }}{a^2}\; = {\text{ }}1/2\;m\;{\omega ^2}{a^2}$


$
  \dfrac{U}{{T.E}} = \dfrac{{\dfrac{1}{2}m{\omega ^2}{x^2}}}{{\dfrac{1}{2}m{\omega ^2}{a^2}}} \\
  \dfrac{U}{{T.E}} = \dfrac{{{x^2}}}{{{a^2}}} \\
  \dfrac{U}{{T.E}} = \dfrac{{{{(\dfrac{a}{2})}^2}}}{{{a^2}}} \\
 $ (since $x$ is half of the amplitude)
$ \Rightarrow \dfrac{{2.5}}{{T.E}} = \dfrac{1}{4}$
By solving;
$T.E = 10J$

Hence the correct answer is Option(B).













Note:
First find the force then the displacement of the particle in the simple harmonic motion and then use the value of the force in finding the work done and finally get the potential energy. Then total energy is equal to sum of kinetic and potential energy but the kinetic energy here is zero.