Question

A particle is vibrating in Simple Harmonic Motion with an amplitude of 4 cm. At what displacement from the equilibrium position does it have half potential and half kinetic?
(A) $1cm$
(B) $2cm$
(C) $\sqrt 2 cm$
(D) $2\sqrt 2 cm$

Answer 1

In this question we will know what are the differences between reflection, refraction, and diffraction. Reflection of light occurs when light rays strike a surface and bounce back or reflect off it, refraction is when light rays enter a different medium of different optical density and change direction, or bend and diffraction is a little bit different phenomenon, in this light wave get spread in different directions because of some obstacle in the path.

Complete step by step solution:
Amplitude of the vibrating particle $a = 4cm$
Also, the question states that $KE$ and $PE$ are both half of the total energy.
Thus, we have
$KE = \dfrac{1}{2}TE = PE$
Also, we know that $TE = KE + PE$
Thus, we come to the conclusion that
$\Rightarrow KE = PE$
Further, substituting the values of $KE$ and $PE$ in the above equation, we get

\Rightarrow \dfrac{1}{2}m{\omega ^2}({a^2} - {x^2}) = \dfrac{1}{2}m{\omega ^2}{x^2} \\\ \\\ $$ (where $m$ is the mass of the object, $\omega $ is the angular frequency, $a$ is the amplitude of the object and $x$ is the position of the object) After cancelling out $$\dfrac{1}{2}m{\omega ^2}$$ on both sides, we get

\Rightarrow ({a^2} - {x^2}) = {x^2} \\
\Rightarrow {x^2} = {a^2} - {x^2} \\
\Rightarrow 2{x^2} = {a^2} \\
\Rightarrow {x^2} = \dfrac{{{a^2}}}{2} \\
\Rightarrow x = \sqrt {\dfrac{{{a^2}}}{2}} \\
\Rightarrow x = \dfrac{a}{{\sqrt 2 }} \\

Substituting the value of amplitude $a$ from the question to the above equation, we get

\Rightarrow x = \dfrac{a}{{\sqrt 2 }} \\
\Rightarrow x = \dfrac{4}{{\sqrt 2 }}cm \\
\Rightarrow x = 2\sqrt 2 cm \\

$$**Therefore, option (D) is the correct answer.** **Note:** Normally when we attempt questions like these, we should convert the values given in the question to SI units. But in this question, the options were given in the same unit as the values in the question. Therefore, there is no need to convert the values to SI units.$$