Question

A particle executes simple harmonic motion of amplitude $A$. At what distance from the mean position is its Kinetic Energy equal to its potential energy?
$\left( A \right)\,0.81\,A$
$\left( B \right)\,0.71\,A$
$\left( C \right)\,0.41\,A$
$\left( D \right)\,0.91\,A$

Answer 1

Here we have to use a simple harmonic distance formula where the amplitude of the simple harmonic motion is $A$. With the help of this distance we can calculate velocity of the particle and as given in the problem we equate kinetic energy with potential energy. Then with the help of distance and velocity we can calculate the mean position.

Complete step by step solution:
Distance from position $X$
$X = A\sin \omega t$
Kinetic Energy $KE$
$KE = \dfrac{1}{2}m{v^2}$
Potential Energy $PE$
$PE = \dfrac{1}{2}K{X^2}$
(where$K = {m^2}{\omega ^2}$)
As per the given problem,
A particle executes simple harmonic motion of the amplitude $A$.
And we know that
Distance from position can be represented as
$X = A\sin \omega t \cdot \cdot \cdot \cdot \left( 1 \right)$
Now we need to know the velocity of the particle,
Differentiating equation $\left( 1 \right)$ wrt to $t$, we get,
$\dfrac{{dX}}{{dt}} = \dfrac{{dA\sin \omega t}}{{dt}} \cdot \cdot \cdot \cdot \left( 2 \right)$
Velocity of a particle V will be,
$V = \dfrac{{\text{Change in distance}}}{{\text{Change in time}}}$
$\Rightarrow V = \dfrac{{dX}}{{dt}}$
Now, equation $\left( 2 \right)$ will become
$V = A\omega \cos \omega t \cdot \cdot \cdot \cdot \left( 3 \right)$
We know
$KE = \dfrac{1}{2}m{v^2} \cdot \cdot \cdot \cdot \left( 4 \right)$
Putting equation $\left( 3 \right)$in$\left( 4 \right)$, we get
$KE = \dfrac{1}{2}m{\left( {A\omega \cos \omega t} \right)^2}$
$\Rightarrow KE = \dfrac{1}{2}m{A^2}{\omega ^2}{\cos ^2}\omega t \cdot \cdot \cdot \cdot \left( 5 \right)$
Now we know,
$PE = \dfrac{1}{2}K{X^2}$
$\Rightarrow PE = \dfrac{1}{2}m{\omega ^2}{X^2} \cdot \cdot \cdot \cdot \left( 6 \right)$
Putting equation $\left( 1 \right)$ in $\left( 6 \right)$, we get
$PE = \dfrac{1}{2}m{\omega ^2}{\left( {A\sin \omega t} \right)^2}$
$\Rightarrow PE = \dfrac{1}{2}m{\omega ^2}{A^2}{\sin ^2}\omega t \cdot \cdot \cdot \cdot \left( 7 \right)$
According to the question,
$KE = PE \cdot \cdot \cdot \cdot \left( 8 \right)$
Putting equation $\left( 5 \right)$and$\left( 6 \right)$respectively in equation$\left( 8 \right)$, we get
$\dfrac{1}{2}m{\omega ^2}{A^2}{\cos ^2}\omega t = \dfrac{1}{2}m{\omega ^2}{A^2}{\sin ^2}\omega t$
Cancelling the common terms we get,
${\sin ^2}\omega t = {\cos ^2}\omega t$
$\Rightarrow {\tan ^2}\omega t = 1$
$\Rightarrow \tan \omega t = 1$
$\Rightarrow \omega t = {\tan ^{ - 1}}1$
$\Rightarrow \omega t = \dfrac{\pi }{4}$
Now we have to calculate the mean position using the distance formula of simple harmonic motion
$X = A\sin \omega t$
Now in place of $\omega t$ we will put $\dfrac{\pi }{4}$,
Hence,
$X = A \times \sin \dfrac{\pi }{4}$
$\Rightarrow X = A \times \dfrac{1}{{\sqrt 2 }}$
$\Rightarrow X = 0.71A$
Therefore the correct option is $\left( B \right)$.

Note: We do not get the kinetic energy directly as we don’t know the velocity of the particle so make use of the distance formula. Correctly put the formula of V an X. Make use of V value to find kinetic energy of the particle and X value for the potential energy.