Question

The ratio of kinetic energy to the potential energy of a particle executing SHM at a distance equal to half its amplitude, the distance being measured from its equilibrium position is
$\begin{align}
  & \text{A}\text{. 3:1} \\
 & \text{B}\text{. 4:1} \\
 & \text{C}\text{. 2:1} \\
 & \text{D}\text{. 8:1} \\
\end{align}$

Answer 1

Hint: A particle is performing simple harmonic motion from its equilibrium position i.e. from mean position. Use the formula of kinetic energy which will give a relation between amplitude, mass and distance. Similarly use formula of potential energy which will give relation between amplitude, mass and distance. Take the ratio, put values and calculate answers.

Complete step by step answer:
A particle is given to be executing SHM at a distance which is equal to half its amplitude.
Let $x$ be the distance and $A$ be the amplitude then mathematically
$x=\dfrac{A}{2}$
Now we have the position of the particle which is $x=\dfrac{A}{2}$.
We know that instantaneous velocity of particles performing S.H.M. at a distance of $x$ from its mean position is
$\begin{align}
  & v=\omega \sqrt{{{A}^{2}}-{{x}^{2}}} \\
 & v=\omega \sqrt{{{A}^{2}}-{{\left( \dfrac{A}{2} \right)}^{2}}} \\
 & {{v}^{2}}=\dfrac{3}{4}A\omega \\
\end{align}$
The expression for kinetic energy of a particle performing S.H.M is given by,
$\begin{align}
  & \text{Kinetic Energy=}\dfrac{1}{2}m{{v}^{2}} \\
 & \text{Kinetic Energy=}\dfrac{1}{2}m\dfrac{3}{4}{{A}^{2}}{{\omega }^{2}} \\
 & \text{Kinetic Energy=}\dfrac{3}{8}m{{A}^{2}}{{\omega }^{2}} \\
\end{align}$
Now we know that term${{\omega }^{2}}$can be written as ${{\omega }^{2}}=\dfrac{k}{m}$
Therefore Kinetic energy can be written as
$\text{Kinetic Energy=}\dfrac{3}{8}k{{A}^{2}}$
The potential energy of a particle performing S.H.M is given by,
$\begin{align}
  & \text{Potential Energy=}\dfrac{1}{2}k{{x}^{2}} \\
 & \text{Potential Energy=}\dfrac{1}{2}k{{\dfrac{A}{4}}^{2}} \\
\end{align}$
Now take the ratio of both kinetic energy and potential energy, we get.
$\begin{align}
  & \dfrac{\text{Kinetic Energy}}{\text{Potential Energy}}=\dfrac{\dfrac{3}{2}k{{\dfrac{A}{4}}^{2}}}{\dfrac{1}{2}k{{\dfrac{A}{4}}^{2}}}=\dfrac{3}{1} \\
 & \dfrac{\text{Kinetic Energy}}{\text{Potential Energy}}=3:1 \\
\end{align}$
The ratio of kinetic energy to the potential energy of a particle executing SHM at a distance equal to half its amplitude, the distance being measured from its equilibrium position is $3:1$

Answer is (A)

Note:
Ratios of similar quantities don’t not possess units. While taking a ratio of two quantities, try to convert their terms in such a way that both quantities will have the same unit. This way it is easy to cancel terms. Using kinetic energy and potential energy, one can calculate the total energy of a particle. Total energy for this specific system is the sum of all the energies present in the system, i.e., kinetic energy and potential energy. Work done by particles can be stored in potential energy.
Velocity of particle is given by, $v=\omega \sqrt{{{A}^{2}}-{{x}^{2}}}$ while acceleration is given by
1. At the mean position, the value of $x$ is zero. Therefore $v=\pm a\omega $ i.e. velocity is maximum i.e. kinetic energy will be more. While at mean position, potential energy will be less.
2. At an extreme position, the value of $x$ is $a$ i.e. $x=a$. Therefore $v=0$ i.e. velocity is minimum i.e. kinetic energy will be less. While at an extreme position potential energy will be more.