Question

A linear simple harmonic oscillator of force constant $2 \times {10^6}N{m^{ - 1}}$ and amplitude $0.01m$has a total energy of $160J$. Its
(A) Maximum potential energy is $100J$
(B) Maximum kinetic energy is $100J$
(C) Maximum potential energy is $160J$
(D) Minimum potential energy is zero.

Answer 1

Hint To solve this question, we need to use the formulae for the kinetic and the potential energy of a linear simple harmonic oscillator. The potential energy is maximum at the extreme position, and minimum at the equilibrium position. The kinetic energy is minimum and maximum at the respective positions.
Formula used: The formulae used for solving this question are given by
$K = \dfrac{1}{2}m{\omega ^2}\left( {{A^2} - {x^2}} \right)$
$U = \dfrac{1}{2}k{x^2}$
Here $K$ and $U$ are the respective kinetic and the potential energies of a simple harmonic oscillator as a function of the displacement $x$ from the mean position, $m$ is the mass, $k$ is the force constant, $\omega $ is the angular frequency, and $A$ is the amplitude.

Complete step-by-step solution:
If the given simple harmonic oscillator oscillates about the mean position of $x = 0$, then we know that the potential energy is given by
$U = \dfrac{1}{2}k{x^2} + {U_m}$.................................(1)
And its kinetic energy is given by
$K = \dfrac{1}{2}m{\omega ^2}\left( {{A^2} - {x^2}} \right)$...........................(2)
Now, we know that the angular frequency is given by
$\omega = \sqrt {\dfrac{k}{m}} $
Squaring both the sides, we get
${\omega ^2} = \dfrac{k}{m}$
Substituting this in (2) we get
$K = \dfrac{1}{2}m\left( {\dfrac{k}{m}} \right)\left( {{A^2} - {x^2}} \right)$
$ \Rightarrow K = \dfrac{1}{2}k\left( {{A^2} - {x^2}} \right)$..............................(3)
Adding (1) and (3), we get
$U + K = \dfrac{1}{2}k{x^2} + \dfrac{1}{2}k\left( {{A^2} - {x^2}} \right) + {U_m}$
$ \Rightarrow U + K = \dfrac{1}{2}k\left( {{x^2} + {A^2} - {x^2}} \right) + {U_m}$
On simplifying we get
$U + K = \dfrac{1}{2}k{A^2} + {U_m}$
We know that the sum of the potential and the kinetic energies is equal to the total mechanical energy. So we get
\[E = \dfrac{1}{2}k{A^2} + {U_m}\]
\[ \Rightarrow {U_m} = \dfrac{1}{2}k{A^2} - E\]
According to the question, $E = 160J$, \[k = 2 \times {10^6}N{m^{ - 1}}\], and $A = 0.01m$. Substituting these in the above equation we get
${U_m} = \dfrac{1}{2}\left( {2 \times {{10}^6}} \right){\left( {0.01} \right)^2} - 160$
$ \Rightarrow {U_m} = \dfrac{1}{2}\left( {2 \times {{10}^6}} \right)\left( {{{10}^{ - 4}}} \right) - 160$
On solving we get
${U_m} = 200 - 160$
$ \Rightarrow {U_m} = 40J$
So the minimum potential energy of the harmonic oscillator is equal to $40J$.
Therefore, the option D is incorrect.
Now, we know that the kinetic energy is maximum at the extreme position, that is, at $x = 0$. Therefore, substituting $x = 0$ in (2), we get
$K = \dfrac{1}{2}k\left( {{A^2} - {0^2}} \right)$
$ \Rightarrow K = \dfrac{1}{2}k{A^2}$
Substituting \[k = 2 \times {10^6}N{m^{ - 1}}\], and $A = 0.01m$ we get
${K_{\max }} = \dfrac{1}{2}\left( {2 \times {{10}^6}} \right){\left( {0.01} \right)^2}$
$ \Rightarrow {K_{\max }} = {10^6} \times {10^{ - 4}}$
On solving, we get
${K_{\max }} = 100J$
So the maximum kinetic energy is equal to $100J$.
Therefore, the option B is correct.
Now, we know that the potential energy is maximum at the extreme positions, that is, at $x = \pm A$ where the kinetic energy is equal to zero. This means that the maximum potential energy is equal to the total energy, which is equal to $160J$.
So the maximum potential energy is equal to $160J$.
Therefore, the option C is also correct and the option A is incorrect.

Hence, the correct options are B and C.

Note: This is a common misconception that the potential energy of a harmonic oscillator at the mean position is zero. But this is not true. It is true only in some special cases. The actual concept is that the potential energy at the mean position is minimum, be it zero or any other value. So the actual formula for the potential energy is $U = \dfrac{1}{2}k{x^2} + {U_m}$ and not $U = \dfrac{1}{2}k{x^2}$.