Question

A particle undergoing simple harmonic motion has time dependent displacement given by $x\left( t \right) = A\sin \left( {\dfrac{{\pi t}}{{90}}} \right)$ . The ratio of kinetic to potential energy of the particle at $t = 210s$ will be:
A. $3$
B. $\dfrac{1}{9}$
C. $\dfrac{1}{3}$
D. $1$

Answer 1

Hint: To solve this question, you can directly apply the formulae of kinetic energy and potential energy at time $t$ when the particle is in simple harmonic motion with an amplitude of $A$ and angular frequency of $\omega $ . Also, remember that the total energy of a particle in simple harmonic motion is: $E = \dfrac{1}{2}m{\omega ^2}{A^2}$ and the sum of kinetic energy and potential energy would always be equal to this value at all times.

Complete step by step answer:
As explained in the hint section of the solution, we can directly use the formulae of kinetic energy and potential energy of a particle in simple harmonic motion at any given time $t$ when the simple harmonic motion has an amplitude of $A$ and angular frequency of $\omega $ . Once we find the value of kinetic energy and the potential energy of the particle at the asked time, all we need to do after that is to find the ratio of kinetic energy to the potential energy to reach at the answer.
The formulae of the energies of a particle in simple harmonic motion are given as:
$
  E = \dfrac{1}{2}m{\omega ^2}{A^2} \\
  KE = \dfrac{1}{2}m{\omega ^2}{A^2}{\cos ^2}\omega t \\
  U = \dfrac{1}{2}m{\omega ^2}{A^2}{\sin ^2}\omega t \\
 $
Now, we have the formulae of kinetic energy $\left( {KE} \right)$ and potential energy $\left( U \right)$ which relates the energies with the following quantities:
$A$ is the amplitude of the simple harmonic motion that the particle is representing
$\omega $ is the angular frequency of the simple harmonic motion
$m$ is the mass of the particle
$t$ is the time at which the value of energies is asked for
Now, let us find the ratio of kinetic energy to potential energy as:
$r = \dfrac{{KE}}{U} = \dfrac{{\dfrac{1}{2}m{\omega ^2}{A^2}{{\cos }^2}\omega t}}{{\dfrac{1}{2}m{\omega ^2}{A^2}{{\sin }^2}\omega t}}$
Upon simplifying, we get the value of the ratio as:
$r = \dfrac{{{{\cos }^2}\omega t}}{{{{\sin }^2}\omega t}}$
We already know that
$\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$
Substituting this in the equation, we get:
$r = {\cot ^2}\omega t$
From the given equation of simple harmonic motion, we can see that the value of the angular frequency is given to be:
$\omega = \dfrac{\pi }{{90}}$
The value of time at which the ratio is asked for is: $t = 210s$
Substituting the above-mentioned values, we get:
$
  r = {\cot ^2}\left( {\dfrac{\pi }{{90}} \times 210} \right) \\
  r = {\cot ^2}\left( {\dfrac{{7\pi }}{3}} \right) \\
 $
We can write it as:
$r = {\cot ^2}\left( {2\pi + \dfrac{\pi }{3}} \right)$
We know that
$\cot \left( {2n\pi + \theta } \right) = \cot \theta $
Hence, we get:
$r = {\cot ^2}\left( {\dfrac{\pi }{3}} \right)$
We already know that
$\cot \left( {\dfrac{\pi }{3}} \right) = \dfrac{1}{{\sqrt 3 }}$
Hence, we now get the value of the ratio as:
$
  r = {\left( {\dfrac{1}{{\sqrt 3 }}} \right)^2} \\
\Rightarrow r = \dfrac{1}{3} \\
 $
Hence, we can see that the correct option is option (C) as the value matches with the value of the ratio that we just found out.

Note:A common mistake that students make is that they get confused and put the sine term in kinetic energy and the cosine term in the potential energy. A solution to this confusion is to remember that at the mean position, the kinetic energy of the particle is maximum while the potential energy is zero, also, the angle is also zero. Hence, we can clearly see that the cosine term will always be in the formula of kinetic energy and the sine term in the formula of potential energy.