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Last updated date: 24th Nov 2023
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# A particle executes SHM on a line $8cm$ long. Its K.E and P.E will be equal when its distance from the mean position is A. $4cm$B. $2cm$C. $\sqrt[2]{2}cm$D. $\sqrt 2$

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Hint: SHM stands for Simple Harmonic Motion, which is characterised as a motion in which the restoring force is proportional to the body's displacement from its mean location. This restoring force often moves in the direction of the mean location. The acceleration of a particle in simple harmonic motion is given by $a\left( t \right){\text{ }} = {\text{ }} - {\omega ^{2\;}}x\left( t \right)$ . Here, is the particle's angular velocity.

Complete step-by-step solution:
Simple harmonic motion is an oscillatory motion in which the particle's acceleration is directly proportional to its displacement from the mean position at any stage. It's a form of oscillatory motion that's a little different.
Simple Harmonic Motions (SHM) are oscillatory and periodic in nature, but not all oscillatory motions are SHM. The harmonic motion of all oscillatory motions, the most common of which is simple harmonic motion, is known as oscillatory motion (SHM).
The displacement, velocity, acceleration, and force in this form of oscillatory motion differ (with respect to time) in a way that can be represented by either sine (or) cosine functions, collectively known as sinusoids.
Now, coming to the question;
$K.E = P.E$
$\Rightarrow \dfrac{1}{2}m{\omega ^2}({A^2} - {x^2}) = \dfrac{1}{2}m{\omega ^2}{x^2}$
$\Rightarrow {A^2} - {x^2} = {x^2}$
$\Rightarrow 2{x^2} = {A^2}$
$\therefore x = \dfrac{A}{{\sqrt 2 }}$
$2A = 8$
$A = \dfrac{8}{2} = 4cm$
$x = \dfrac{A}{{\sqrt 2 }}$
$x = \dfrac{4}{{\sqrt 2 }} = \sqrt[2]{2}$
So, the correct option is: (C) $\sqrt[2]{2}cm$.

Note:Simple Harmonic Motion is a very useful and important method for understanding the properties of sound waves, light waves, and alternating currents. Any non-simple harmonic oscillatory motion can be expressed as a superposition of several harmonic oscillatory motions of different frequencies.