Question

Displacement between maximum potential energy position and maximum kinetic energy position for a particle executing simple harmonic motion is.

Answer 1

Hint: According to the law of conservation of energy, the system's overall energy will never change. For straightforward harmonic motion, the displacement of the particle from the mean location determines the exchange of kinetic and potential energy.

Complete step by step solution:
In order to know that the expression of the kinetic energy is:
\[K = \dfrac{1}{2}m{\omega ^2}({a^2} - {x^2})\]..…. (1)
where m is the mass of the body, ω is the angular frequency, a is the amplitude and x is the position of the particle.

And, the expression of potential energy is:
\[U = \dfrac{1}{2}m{\omega ^2}{x^2}\]…… (2)
where m is the mass of the body, ω is the angular frequency, a is the amplitude and x is the position of the particle.

We observe that at the mean position \[(x = 0)\], kinetic energy is maximum \[\left( {\dfrac{1}{2}m{\omega ^2}{a^2}} \right)\] and potential energy is minimum or zero.

Also, at extreme positions \[(x = \pm a)\], kinetic energy is zero and potential energy is maximum \[\left( {\dfrac{1}{2}m{\omega ^2}{a^2}} \right)\].

Thus, the displacement between positions of maximum potential energy and maximum kinetic energy is \[ \pm a\].

Note:Here, it's important to keep in mind that the system's overall energy should not change. The answer will then be aided by the fact that the energy of the Simple Harmonic Oscillator only totally transforms from one form to another.