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Physics Bootcamp

Section 13.3 Simple Harmonic Systems

Simple harmonic motion appears in a wide variety of physical settings. The defining characteristic of a system exhibiting a simple harmonic motion in some variable \(x\) is that second derivative of the variable is prportional to the negative of \(x\text{.}\)
\begin{equation*} \frac{d^2x}{dt^2} = - \omega^2 x. \end{equation*}
Variable \(x\) does not have to be \(x\)-coordinate. It can be some angle, e.g., in a simple pendulum. In the following sections we will cover first three such systems. The other systems will be studied in future chapters.
  1. Plane Pendulum
  2. Physical Pendulum
  3. Torsion Pendulum
  4. Oscillations of Floating Objects
  5. Electromagnetic Oscillations in LC Circuits
When we study energy of simple harmonic oscillators, we will find another way of identifying systems that exhibit a simple harmonic motion. Briefly, their energy will sume of two quadratice terms, one coming from the kinetic energy and the other from the potential energy. Let \(v\) be the velocity (i.e. \(v = dx/dt\) ), then, energy will be
\begin{equation*} E = \frac{1}{2}\,m\,v^2 + \frac{1}{2}\,k\,x^2. \end{equation*}
The ratio of the coeficients of \(x^2\) and \(v^2\) give the square of the angular frequecy of the oscillations.
\begin{equation*} \omega^2 = \frac{ \frac{1}{2}\,k }{ \frac{1}{2}\,m } = \frac{k}{m}. \end{equation*}

Remark 13.17. General Applicablity of Simple Harmonic Motion.

Simple harmonic can be found near potential energy minima which does not have zero second derivative, e.g., near \(x=x_0\) and \(x+x_1\) in Figure 13.18. Thus, near \(x=x_0\text{,}\) potential energy \(U(x)\) can be approximated by using the Taylor expansion about \(x=x_0\) to give a quadratic approximation (A similar expansion about \(x=x_1\) will give another approximation with a different \(k\)).
\begin{align*} U(x) \amp = U(x_0) + \frac{1}{2}k\,(x - x_0)^2 + \cdots \end{align*}
where \(\left.\frac{dU}{dx}\right|_{x=x_0} = 0\) and \(k = \left.\frac{d^2 U}{dx^2}\right|_{x=x_0}\,(x - x_0) \gt 0\) since minimum there and we assume that second derivative does not vanish at this minimum.
Near these minima (here \(x=x_0\) and \(x=x_1\) ), the system will have a restoring force that pushes the system back toward the equilibrium with a linear force since force is negative of the potential energy
\begin{equation*} F = -\frac{dU}{dx} = - k x. \end{equation*}
Figure 13.18.
Therefore, every physical system with these characteristics will act like a simple harmonic system with a characteristic frequency.

Subsection 13.3.1 Motion Near Potential Minima

From our discussion in this chapter, you know that that a restoring force that is proportional to the displacement from the equilibrium and points in the opposite direction will lead to a Simple Harmonic Motion. Now, the \(x\)-component of a conservative force \(\vec F\) is related to potential energy \(U\) as follows,
\begin{equation*} F_x = -\frac{dU}{dx}. \end{equation*}
Therefore, any potential energy that is quadratic in \(x\text{,}\) the displacement variable, will result in the restoring force appropriate for a Simple Harmonic Motion. This is obviously the case with the potential energy due to force from an ideal spring.
In general, consider a potential energy \(U(x)\) that has a minimum at \(x=x_0\text{.}\) By writing the potential energy function in terms of a Taylor series about \(x=x_0\) we obtain the following.
\begin{equation*} U(x)= U(x_0) + \left(\frac{dU}{dx}\right)_{x=x_0}\left(x-x_0\right) + \frac{1}{2!}\left(\frac{d^2U}{dx^2}\right)_{x=x_0}\left(x-x_0\right)^2 + \cdots \end{equation*}
Since the potential energy has a minimum at \(x=x_0\text{,}\) the first derivative is zero there, and the leading non-constant term is the quadratic term in \(x-x_0\text{,}\) the displacement from the equilibrium.
\begin{equation*} U(x)= U(x_0) + \frac{1}{2!}\left(\frac{d^2U}{dx^2}\right)_{x=x_0}\left(x-x_0\right)^2 + \cdots \end{equation*}
The value of the second derivative of the potential energy function for \(x=x_0\) is a constant. Let us denote this constant by \(k\) in anticipation of its analogy with the spring constant of a spring.
\begin{equation*} k \equiv \left(\frac{d^2U}{dx^2}\right)_{x=x_0}. \end{equation*}
Choosing the potential energy to be zero at the equilibrium, and placing the origin at the equilibrium point, we find that near a potential energy minimum, the leading behavior of the potential energy function is quadratic.
\begin{equation*} U(x)= \frac{1}{2!}kx^2+\cdots \end{equation*}
Therefore, even though an oscillating system may not be a block attached to a spring, the behavior is “identical” to the problem of a block attached to a spring and we can speak of a “spring constant” whenever a system is oscillating such that near the bottom of the potential energy the potential energy can be approximated by a quadratic function of the corresponding displacement. The only exceptions are those potential energy functions, such as \(U(x) = b x^4\text{,}\) which cannot be approximated by a quadratic function near the minima.
A quadratic potential energy function gives a linear restoring force of the Hooke’s law and leads to the Simple Harmonic Motion.
\begin{equation*} F_x= -\frac{dU}{dx}= -k x + (\textrm{ higher powers in }x). \end{equation*}
We have seen above that the plane pendulum is not a Simple Harmonic Oscillator unless the angle of oscillation is small. We can see this emerging Simple Harmonic property from the perspective of a quadratic potential energy function. The potential energy of a pendulum when it is displaced at angle \(\theta\) is
\begin{equation*} U = m\, g\, l\, (1- \cos \theta). \end{equation*}
Now, expanding \(\cos (\theta)\) for small \(\theta\) we find
\begin{equation*} \cos(\theta) = 1-\frac{1}{2!}\theta^2 + \frac{1}{4!}\theta^4 + .... \end{equation*}
If we keep only the leading term, viz., \(1\text{,}\) we will lose all physics information associated with \(\theta\text{.}\) Therefore, we will keep two terms in this expansion. This gives the following expression for the potential energy near \(\theta=0\text{:}\)
\begin{equation*} U = \frac{m g l}{2}\theta^2, \end{equation*}
which is quadratic in the dynamical variable \(\theta\text{.}\) Hence, for small angles we expect a Simple Harmonic Motion for the pendulum as discussed previously.

Remark 13.19. Force Vector from Potential Energy Function.

For, a general case, we need to use partial derivatives.
\begin{align*} \vec F \amp = -\left( \frac{\partial U}{\partial x}\hat u_x + \frac{\partial U}{\partial y}\hat u_y\right. + \left.\frac{\partial U}{\partial z}\hat u_z \right), \end{align*}
where \(\hat u_x\text{,}\) \(\hat u_y\text{,}\) and \(\hat u_z\) are unit vectors pointed towards the positive \(x\text{,}\) \(y\) and \(z\)-axes respectively.

Example 13.20. Simple Harmonic Motion About a Minimum of a Cubic Potential.

The potential energy of a particle of mass \(m\) moving along \(x\)-axis is given as \(U(x)=x(x-1)^2\text{.}\)
  1. Plot the potential energy as a function of \(x\text{.}\) Discuss the types of motion that a particle will execute for various values of energy of the particle.
  2. Find the location of the minimum of the potential energy function.
  3. Find the angular frequency of small oscillation about the minimum of the potential.
  4. Are there any restrictions on the displacement for the particle to oscillate about the minimum? Explain.
Solution 1. a,b
Figure 13.21 presents the plot. The plot shows a minumum at \(x=1\) and a maximum at \(x=1/3\text{.}\)
Figure 13.21.
If the energy of the particle is between 0 and \(\frac{4}{27}\) and \(0.33 \lt x \lt 1.33\) then the particle will oscillate about \(x=1\text{.}\) If the energy is larger than \(\frac{4}{27}\text{,}\) then no matter where the particle is released, the particle will go towards \(x=-\infty\text{.}\)
Solution 2. c
For small oscillations about \(x=1\) we can express \(U(x)\) about \(x=1\) to determine the corresponding \(k\) of the restoring force. Let \(\xi = (x-1)\text{.}\) The potential energy about \(x-1\) is obtained by doing a Taylor series about \(x=1\text{.}\)
\begin{equation*} U (\xi) = \xi^2 \end{equation*}
In analogy with potential energy of a simple harmonic oscillator, \(\frac{1}{2} kx^2\) we conclude that \(k=2\) here. Therefore, the angular frequency of the particle for the oscillations about \(x=1\) will be
\begin{equation*} \omega = \sqrt{\frac{2}{m}}. \end{equation*}
Solution 3. d
The oscillations will occur only when the energy is such that the particle will be trapped in the well around \(x=1\text{.}\)

Exercises 13.3.2 Exercises

1. Simple Harmonic Motion About a Minimum of a Quartic Potential.

The potential energy of a particle of mass \(m\) moving along \(x\)-axis is given as \(U(x)=x^2(x-1)^2\text{.}\)
  1. Plot the potential energy as a function of \(x\text{.}\) Discuss the types of motion that a particle will execute for various values of energy of the particle.
  2. Find the location of the minimum of the potential energy function.
  3. Find the angular frequency of small oscillation about the minimum of the potential.
  4. Are there any restrictions on the displacement for the particle to oscillate about the minimum? Explain.
Hint.
The problem is similar to Example 13.20. There are two wells now in which the particle can oscillate with simple harmonic motions. Follow the same procedure presented there to discuss these oscillations.