Question

The total energy of a particle executing Simple harmonic motion S.H.M. is proportional to the following:

(A) Displacement from equilibrium position of particle
(B) Frequency of oscillation
(C) Velocity in equilibrium position
(D) Square of amplitude of motion

Answer 1

Hint:
First start with finding the relation of 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 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 :
$T.E. = \dfrac{1}{2}m{\omega ^2}{a^2}$



Complete step by step solution:

The total energy of a particle executing in a simple harmonic motion is the sum of the potential energy and kinetic energy of the particle in the simple harmonic motion (S.H.M.).
Let the total energy be T.E. = K.E. + P.E.
Kinetic energy be $K.E. = \dfrac{1}{2}K\left( {{a^2} - {x^2}} \right)$
Potential energy be $P.E. = \dfrac{1}{2}K{x^2}$
So, $T.E. = \dfrac{1}{2}K\left( {{a^2} - {x^2}} \right) + \dfrac{1}{2}K{x^2}$
By solving;
$T.E. = \dfrac{1}{2}K{a^2}$
Where, $K = m{\omega ^2}$
Therefore, $T.E. = \dfrac{1}{2}m{\omega ^2}{a^2}$
So,
$T.E. \propto {a^2}$
Total energy is directly proportional to amplitude of motion.

Hence the correct answer is Option(D).









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.