Question

The total energy of a particle executing SHM is $80\,J$ . What is the kinetic energy of the particle when it is at a distance of $\dfrac{3}{4}$ of amplitude from the mean positive?
A. $45\,J$
B. $35\,J$
C. $55\,J$
D. $25\,J$

Answer 1

To solve the question, we will use the total energy formula to calculate the particle's kinetic energy while it is at a particular amplitude from the mean positive and also because the system is instantaneously at rest the value of $K$ is zero.

Complete step by step answer:
In the given question it is given to us that the total energy is $80\,J$. And we know that, if a conservative force applies on an oscillating particle, its total energy $\left( E \right)$ is the sum of its kinetic and potential energy.
Total energy = Kinetic energy + Potential energy
And kinetic energy is equal to zero. So,
Total energy=$\dfrac{1}{2}m{{\omega}^2}{A^2}=\dfrac{1}{2}k{A^2}$
Where, $A$ is for amplitude, and $k$ is for a positive constant.
$\Rightarrow 80 = \dfrac{1}{2}k{A^2}$

When the particle is $\dfrac{3}{4}$ of an amplitude away from the mean position, we must calculate the potential energy.Assume the particle is at a distance of $'x'$ from the mean location. So,
Potential energy=$\dfrac{1}{2}k {x^2}$
And, in accordance with the given circumstances or in response to the query,
$x{\text{ }} = {\text{ }}\dfrac{3}{4}A$
So,
$\text{Potential energy} = {\text{ }}\dfrac{1}{2} \times {\text{ }}k{\text{ }} \times {\text{ }}{\left( {\dfrac{{3A}}{4}} \right)^2}$
$\Rightarrow \text{Potential energy} =\dfrac{1}{2} \times k \times \dfrac{{9{A^2}}}{{16}} \\\ \Rightarrow \text{Potential energy} =\dfrac{9}{{16}} \times \dfrac{{k{A^2}}}{2} \\\ \Rightarrow \text{Potential energy} =\dfrac{9}{{16}} \times 80\,\,\,\,\,\,\left[ {\because k{A^2} = 80} \right]$
After the final evaluation we will get our answer as
$\therefore \text{Potential energy} =45J$

So, the correct option is A.

Note: When a particle is displaced from its mean location in simple harmonic motion, its kinetic energy is transformed to potential energy and vice versa, while total energy remains constant. Simple harmonic motion's total energy is independent of $x$.