Energy of a Harmonic Oscillator

Section 13.7 Energy of a Harmonic Oscillator

Spring force is a conservative force with the following potential energy formula.

\begin{equation} U = \dfrac{1}{2} k x^2,\tag{13.39} \end{equation}

where the zero potential energy reference corresponds to the situation when spring is in its original relaxed state. By adding this to the kinetic energy of the block we get the energy \(E\) of the block. Let us write energy at time \(t \) as \(E(t) \text{.}\)

\begin{equation} E(t) = \dfrac{1}{2}m v^2 + \dfrac{1}{2} k x^2,\tag{13.40} \end{equation}

where for brevity, I have dropped the subscript \(x\) from \(v_x \text{.}\) By plugging in the solution \(x(t)= A\cos(\omega t + \phi)\) and \(v(t)= -\omega A\sin(\omega t + \phi)\text{,}\) and using \(k = m\omega^2\text{,}\) you can show that

\begin{equation} E(t) = \dfrac{1}{2} k A^2,\tag{13.41} \end{equation}

or, equivalently,

\begin{equation} E(t) = \dfrac{1}{2} m v_0^2 + \dfrac{1}{2} k x_0^2,\tag{13.42} \end{equation}

which is just the energy at the initial time. That is, the energy of simple harmonic oscillator is conserved.

\begin{equation} E(t) = E(0).\tag{13.43} \end{equation}

Although, the sum of kinetic and potential energies remains unchanged, kinetic and potential energies change with time. The potential energy is maximum when the oscillator is at the turning points of the motion, where velocity is zero, and hence kinetic energy is zero. You might say that the block spends most of its time near the two ends of its motion, where it is moving the slowest.

When the block is at \(x = 0 \text{,}\) potential energy is zero, which is the minimum value of potential energy. That means kinetic energy is largest at \(x=0 \text{,}\) i.e, when the block is at the equilibrium position it moves the fastest.

Example 13.25. Energy Stored in a Spring.

(a) A spring of length \(110\text{ cm}\) and spring constant \(1.5\times 10^{4}\text{ N/m}\) is compressed such that its length shrinks by \(20\text{ cm}\text{.}\) What is the amount of energy stored in the spring?

(b)The same spring is strethed such that its length expands by \(20\text{ cm}\text{.}\) What is the amount of energy stored in the spring?

Answer.

(a) \(300\text{ J}\text{,}\) (b) \(300\text{ J}\text{.}\)

Solution.

(a) The change in length \(\Delta l\) is what we need, not the length of the spring. The energy stored is

\begin{equation*} U =\dfrac{1}{2}k (\Delta l)^2 = \dfrac{1}{2}\times 1.5\times 10^{4} \times 0.20^2 = 300\text{ J}. \end{equation*}

(b) The energy in spring is same for compression and expansion, since the energy depends on the square of the change.

\begin{equation*} U = 300\text{ J}. \end{equation*}

Example 13.26. Speed of a Block Attached to a Spring.

A block of mass \(0.6\text{ kg}\) is attached to a spring of spring constant \(180\text{ N/m}\) and negligible mass compared to the mass of the block. The block is placed on a frictionless horizontal table, then pulled horizontally \(3\text{ cm}\) from its equilibrium position and let go from rest. Evaluate speed of the block (a) as it crosses the equilibrium position, and (b) when it is \(1.5\text{ cm}\) from the equilibrium.

Answer.

(a) \(0.52\text{ m/s}\text{,}\) (b) \(0.45\text{ m/s}\text{.}\)

Solution.

Since block has no friction on it, its energy will be conserved. Let us denote the original position as point A, equilibrium as point B and \(1.5\text{ cm}\) from the equilibrium as point C.

(a) When block is at point A, it is not moving, therefore all energy is in the potential energy stored in the spring. We convert all length units to \(\text{meter}\text{.}\)

\begin{equation*} E_A = \frac{1}{2}kx_A^2 = \frac{1}{2}(180)(0.03)^2 = 0.081\text{ J}. \end{equation*}

When block is at point B, the spring is neither stretched nor compressed, therefore there is no potential energy stored in the spring. Hence, all the energy which the system started out with must be in the form of kinetic energy of the block.

\begin{equation*} E_B = \frac{1}{2}mv_B^2 = \frac{1}{2}(0.6)\,v_B^2 = 0.3\, v_B^2. \end{equation*}

Equating the energy at B to the energy at A we find

\begin{equation*} 0.3\, v_B^2 = 0.081\ \ \Longrightarrow\ \ v_B = 0.52\text{ m/s}. \end{equation*}

(b) When block is at point C, which could be on either side of the equilibrium a distance of \(1.5\text{ cm}\) away from the equilibrium, the spring is either compressed or stretched. Therfore, there will be potential energy stored in the spring. But, the stored potential energy is less than the starting energy, hence, the rest of the energy will be in the form of kinetic energy of the block.

\begin{equation*} E_C = \frac{1}{2}mv_C^2 + \frac{1}{2}kx_C^2= 0.3\, v_C^2 + 0.020. \end{equation*}

Equating \(E_C\) to \(E_A\text{,}\) and solving for \(v_C\) we find the speed when the block is \(1.5\text{ cm}\) from the equilibrium to be

\begin{equation*} 0.3\, v_C^2 + 0.020 = 0.081 \ \ \Longrightarrow\ \ v_C = 0.45\text{ m/s}. \end{equation*}

Subsection 13.7.1 Potential Energy and Turning Points

Often we plot a potential energy diagram to display the turning motions at a particular energy value. Figure 13.27 shows one such plot of potential energy versus position. The horizontal energy line on the diagram reflects the fact that energy is same at any position taken by the oscillator. The oscillator moves between the turning points, where velocity turns around in direction. At these points velocity must be zero - energy at these points are all in potential energy. When \(x = 0\) all energy in kinetic energy. At other values of \(x\text{,}\) the energy has both kinetic and potential energies.

Figure 13.27. The energy diagram shows visually, the conversion between kinetic and potential energies in a simple harmonic motion, keeping the total energy fixed.

Exercises 13.7.2 Exercises

1. Energy of a Simple Harmonic Oscillator at Different Points of the Cycle.

Consider a simple harmonic oscillator of mass \(m \text{,}\) amplitude \(A\) and frequency \(f\text{.}\)

(a) What fraction of the energy of the oscillator is in the kinetic energy when the oscillator’s displacement is half the amplitude?

(b) What fraction of the energy of the oscillator is in the potential energy when the oscillator’s displacement is half the amplitude?

(d) Where in the cycle of an oscillation does the oscillator have the largest speed? Why?

Hint.

(a) and (b) Use conservation of energy, (c) Think what happens at the turning points, (d) Think when the PE is lowest, i.e., zero.

Answer.

(a) \(75\% \text{,}\) (b) \(25\% \text{,}\) (c) Turning points, \(x=\pm A \text{,}\) (d) At equilibrium point, \(x=0\text{,}\) \(v = 2\pi f A\text{.}\)

Solution 1. (a),(b)

(a) From energy conservation, we know that energy at any time in the cylce must equal the energy at the turning points. At turning points, energy is all potential energy.

\begin{equation*} E = \dfrac{1}{2}kA^2, \end{equation*}

where

\begin{equation*} k = m\omega^2 = 4\pi^2 m f^2. \end{equation*}

When oscillator’s displacement is half of \(A \text{,}\) we will get

\begin{equation*} KE + \dfrac{1}{2} k \left( \dfrac{A}{2}\right)^2. \end{equation*}

Equating the two, we get

\begin{equation*} KE = \dfrac{1}{2} k A^2 \times \dfrac{3}{4}. \end{equation*}

Therefore,

\begin{equation*} \dfrac{KE}{E} = \dfrac{3}{4}. \end{equation*}

(b) Potential Energy at this instant is \(25\% \) of the total energy.

Solution 2. (c),(d)

(c) Lowest speed is zero which occurs at the turning points, when the displacment has values, \(x = -A\) and when \(x = A\text{.}\)

(d) Largest speed will occur where KE is largest, which would occur where the PE is the least. The least PE is zero here and occurs when \(x = 0\text{.}\) Therefore, at equilibrium position

\begin{equation*} \dfrac{1}{2}mv^2 = \dfrac{1}{2} k A^2. \end{equation*}

Therefore, maximum speed is

\begin{equation*} v = \sqrt{\dfrac{k}{m}}\ A = \omega\, A = 2\pi f A. \end{equation*}

2. Analyzing Energy and Motion of a Block Hung from the Ceiling.

A block of mass \(5\, \text{kg}\) is attached to a spring of the original length is \(50\) cm and the spring constant \(100\) N/cm. Other end of the spring is fixed to a rigid support in the ceiling, which is \(2\, \text{m}\) above the floor and the block is gently hung. As a result of the weight of the block, the spring stretches to a different length at equilibrium. The block is then hit with a hammer giving it an instantaneous velocity of \(4\) cm/s downward.

When the block is hanging in equilibrium, where is the equilibrium position of the block with respect to the floor?
Pick the zero reference points for the potential energies due to the spring force and that due to the gravity, and determine the potential energy of the block at the time it was hit by the hammer?
What is the kinetic energy of the block immediately after being hit by the hammer?
How much work did the hammer do?
What is the frequency of the oscillations of the block?
What is the phase constant of the motion of the block?
Suppose the \(y\)-axis is pointed up with the origin at the equilibrium when the block is hanging with zero net force on the block, write a function \(y(t)\) for the \(y\)-coordinate of the block.
Find the turning points in the motion of the block.

Solution 1. a

When the block is hanging and the block is not oscillating, the forces on the block will be balanced. This gives the stretch \(\Delta l\) to be

\begin{equation*} \Delta l = \frac{mg}{k} = \frac{5\ \textrm{kg}\times 9.81\ \textrm{m/s}^2}{10000\ \textrm{N/m}} = 0.391\ \textrm{cm}. \end{equation*}

Therefore, the block will be 200 cm - (50.39) cm = 149.61 cm from the floor.

Solution 2. b

Let the point 149.61 cm from the floor be the reference zero for both potential energy due to the spring force and due to the gravity. Then, the net potential energy the block at that point will be zero.

Solution 3. c

The kinetic energy will be

\begin{equation*} \textrm{KE}\ = \frac{1}{2} \times 5 \textrm{kg} \times (0.04\ \textrm{m/s})^2 = 0.004 \ \textrm{J}. \end{equation*}

Solution 4. d

The work by hammer is equal to the change in the kinetic energy of the block here. Therefore, the work by hammer = 0.004 J.

Solution 5. e

The frequency of the oscillations of the block depends on the mass and the spring constant.

\begin{equation*} f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} = \frac{1}{2\pi} \sqrt{\frac{10000\ \textrm{N/m}}{5\ \textrm{kg}}} = 7.12\ \textrm{Hz}. \end{equation*}

Solution 6. f

The hammer is assumed to be in contact for a brief moment. After that the block oscillates under the influences of the gravity and the spring force. The equilibrium position of the block for these forces was at 149.61 cm from the floor. Let the origin of the coordinate system be placed at the point and let us point the \(y\)-axis up. With this choice we have the following initial conditions for a simple harmonic motion about \(y=0\text{.}\)

\begin{equation*} y(0) = 0; v_{y} (0) = - 0.04\ \textrm{m/s}. \end{equation*}

If the phase constant starts at zero at maximum positive \(y\text{,}\) then the motion with \(y=0\) and block moving towards the negative \(y\) axis will be ahead of the motion of zero phase by \(pi/2\) radians, i.e. quarter of a cycle.

Solution 7. g,h

From the phase description we might expect that

\begin{equation*} y(t) = A \cos(\omega t +\pi/2), \end{equation*}

with \(\omega = \sqrt{k/m}\text{.}\) This simple answer turns out to be misleading. This answer says that the box will come to rest at equal distance on either side of \(y=0\text{.}\) However, a consideration of the energy conservation shows that the movement of the block above \(y=0\) is not equal to the movement below \(y=0\text{.}\)

Let \(y=y_0\) be the coordinate of the place where the tip of the spring lies when not attached to the block. Let \(y=A_1\) when the block is above \(y=0\) when coming to rest and \(y=-A_2\) when the block is below \(y=0\) and comes to rest. Denoting the energy of the block when it is at \(y=0\text{,}\) \(y=A_1\text{,}\) and \(y=-A_2\) by \(E_0\text{,}\) \(E_1\text{,}\) and \(E_2\) respectively. The energy of the block at these points will be

\begin{align*} \amp E_0 = \frac{1}{2} m v_0^2.\\ \amp E_1 = \frac{1}{2} k A_1^2 + m g A_1.\\ \amp E_2 = \frac{1}{2} k A_2^2 - m g A_2. \end{align*}

Notice that the energy conservation requires that these must be equal. Equating them does not guarantee \(A_1 = A_2\text{.}\) Equating \(E_1\) to \(E_2\) gives

\begin{equation*} \frac{1}{2} k A_1^2 + m g A_1 = \frac{1}{2} k A_2^2 - m g A_2 \end{equation*}

This shows that the block will go down a little further from \(y=0\) than it goes up. The difference

\begin{equation*} A_2 - A_1 = \frac{2 mg}{k}. \end{equation*}

For stiff springs, the difference will be negligible and the block will oscillate as the cosine solution given above. In the present case,

\begin{equation*} A_2 - A_1 = \frac{2 mg}{k} = 1\textrm {cm}. \end{equation*}

From equating \(E_2\) to \(E_0\) we find \(A_2 \approx 1\) cm. Therefore, in our case we cannot ignore the difference. The numerical values in this problem are not very good. They do not illuminate the importance of energy conservation and turning points.

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