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If $ < E > $ and $ < U > $ denotes the average kinetic and average potential energies respectively of mass describing a simple harmonic motion over harmonic motion over one period then the correct relation is :
A) $ < E > = < U > $
B) $ < E > = 2 < U > $
C) $ < E > = - 2 < U > $
D) $ < E > = - < U > $




Answer
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Hint:
To solve this question we must know the formulas of kinetic and potential energy in SHM. By using these formulas we can find the relationship between them and solve the question.


Complete step by step solution:

In the question, we have given the average kinetic energy and the average potential energies as $ < E > $and $ < U > $respectively,

The particle's kinetic energy with mass $m$, equals:
$K = \dfrac{1}{2}m{\omega ^2}({a^2} - {x^2})$
While 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.
$U = \dfrac{1}{2}m{\omega ^2}{x^2}$
As we know that the total energy is sum of the kinetic and potential energy, then we have:
 $E = \dfrac{1}{2}m{\omega ^2}({a^2} - {x^2}) + \dfrac{1}{2}m{\omega ^2}{x^2} \\$
$E = \dfrac{1}{2}m{\omega ^2}{a^2} \\$
For a whole cycle in simple harmonic motion, the average values of kinetic and potential energy are equal and a particle that is moving in the simple harmonic motion which has a total energy of $\dfrac{1}{2}m{\omega ^2}{a^2}$.
$\therefore < E > = < U > = \dfrac{1}{2}m{\omega ^2}{a^2}$
Thus, the correct option is: (A) $ < E > = < U > $





Note:
It should be noted that the kinetic energy and the potential energy are equivalent to each other. The total energy in the simple harmonic motion which is entirely kinetic in the middle and entirely potential in the extreme positions. And an energy in the simple harmonic motion which is an entire amount of the energy that is a particle which holds when engaging in simple harmonic motion.