Kinetic Energy as a
Function of Time
Prerequisites
Students should be familiar with the general properties of simple harmonic motion (SHM), in particular, the way in which the displacement and the velocity of the oscillating object depend on time. Students should be familiar with the concepts of kinetic and potential energy and energy conservation.
Learning Outcomes
Students will learn how to use energy conservation to obtain information about the potential energy of a system consisting of a vertical spring, a block suspended from it, and the earth. They will be able to describe the time dependence of the potential energy when the block is oscillating and to explain how the potential energy depends on the displacement of the weight from its equilibrium position. They will learn that the energy of the spring-block-earth system is proportional to the square of the amplitude of oscillation of the system.
Instructions
Students should know how the applet functions, as described in Help and ShowMe.
The applet should be open. The step-by-step instructions in the following text are to be done in the applet. You may need to toggle back and forth between instructions and applet if your screen space is limited.
Kinetic Energy as a
Function of Time
Potential Energy as a
Function of Time
Potential Energy as a Function
of Position
Energy as a Function of
Amplitude
The kinetic energy KE of weighted spring varies in time when the block is oscillating up and down. Let's find out how KE depends on t.
Exercise 1. RESET 
the applet.
Set the spring constant to k = 200 N/m, and keep the mass and the amplitude at their default values of m = 1.25 kg and A = 0.20 m, respectively.
PLAY the motion from its default position, which has the block in its top position, and observe the block's motion.
Based on your qualitative observation of the block's motion, sketch graphs of position vs. time (y vs. t) and velocity vs. time (vy vs. t) for the block. y is the position of the center of the block. The y-axis increases in the upward direction, with y = 0 at the block's equilibrium position. On the graphs, mark the points where the block passes its top, bottom, and equilibrium positions.
When done, display the applet's y vs. t and vy vs. t graphs to check your graphs.
Exercise 2. Continuing from Exercise 1, REWIND the applet and PLAY the motion. Observe the size of the kinetic energy column during the motion, and, based on this qualitative observation, sketch the kinetic energy KE of the oscillating block vs. t. You may find it helpful to STEP through the motion.
On the graph, mark the points where the block passes its top, bottom, and equilibrium positions. Make sure your graph is consistent with the vy vs. t graph obtained in Exercise 1. When done, display the applet's KE vs. t graph to check your graph.
Exercise 3. Write down general expressions in symbols for the block's position y(t) and velocity vy(t) at time t for the motion played in Exercises 1 and 2 in terms of the amplitude A, angular frequency w, and time t. If you have difficulty with this Exercise, review the Lesson "Simple Harmonic Motion (Vertical Spring)" accompanying the applet "Simple Harmonic Motion".
Using the expression for vy(t), write down an expression for KE(t) in terms of A, w, and the mass m, and, of course, t.
Answer. For the kinetic energy, you should obtain
KE(t) =
(m/2)A2w2 sin2
wt .(1)
Exercise 4. REWIND the applet, and STEP through the motion until the block is as close to its equilibrium point as possible (the energy column would be entirely red at the equilibrium point). Record the value of the kinetic energy at this point from the Data box. Calculate this value using Equ.(1) above and compare your result to the value of KE in the Data box.
Calculate the kinetic energy at another instant during the block's motion. Also, click Rewind, change some of system parameters, and Step the system to a time of your choice. Calculate the kinetic energy at this time from Equ.(1). Compare your values with those displayed in the Data box.
The mechanical energy E, energy for short, of the spring-block-earth system stays constant during the system's motion (if the motion is undamped, as it is in the applet). One says the energy is conserved. The energy is the sum of kinetic and potential energy,
E = KE + PE. (2)
Play the motion and observe how there is an ongoing conversion of potential into kinetic energy or vice versa during the motion such that the sum of the two energies remains constant.
In this section, you will use Equ.(2) together with Equ.(1) to obtain information about the time dependence of the potential energy.
Exercise 1. RESET the applet. Set the spring constant to k = 200 N/m, and keep the mass and the amplitude at their default values of m = 1.25 kg and A = 0.20 m, respectively. These are the same settings as at the beginning of the preceding section.
PLAY the motion, and sketch a graph of the potential energy as function of time. Mark the points where the block passes its top, bottom, and equilibrium positions. When done, compare your sketch to the graph plotted by the applet.
Exercise 2. Continuing from Exercise 1, determine the value of the potential energy at the block's top and bottom positions and the value of the mechanical energy as follows.
The potential energy of any system is defined up to an additive constant whose value one can choose freely to suit one's convenience. In the applet, as is commonly done for the present system, this constant is chosen so that the potential energy of the spring-block-earth system is 0 when the block is at its equilibrium position. Therefore, at this point, the mechanical energy E is equal to the kinetic energy KE.
When the block is either at its top or bottom positions, the block is momentarily at rest and therefore its kinetic energy equal to zero. Thus, Equ.(2) implies that E = PE at this point.
Therefore, since E is constant during the motion, the potential energy at the top or bottom is equal to the kinetic energy when the block is passing its equilibrium position, which is the maximum kinetic energy:
PEtop,bottom = E =
KEmax. (3)
Use Equ.(3) combined with Expression (1) for the kinetic energy, to derive a general expression for the potential energy at the top or bottom of the motion and for the energy E. Hint: what is the maximum value of sin2 wt?
Answer. Since the maximum value of the sine function and that of its square are equal to 1, Equ.(1) implies
PEtop,bottom = E =
(m/2)A2w2. (4)
Use Equ.(4) to calculate the values of PEtop,bottom and E for the present applet settings. Compare your results to the values in the Data box and also with the value of the kinetic energy near the equilibrium point determined in Exercise 4 of the previous section. The latter may not be quite equal to KEmax if you stopped the motion not exactly at the equilbrium point.
Exercise 3. In the Lesson "Simple Harmonic Motion (Vertical Spring)" accompanying the applet "Simple Harmonic Motion" it is shown that the angular frequency w is related to the spring constant k and the mass m of the block by
w2 =
k / m .(5)
Substitute this expression for w2 into Equ.(4) to obtain an expression for the energy in terms of k and A.
The result is
PEtop,bottom = E =
(k/2)A2 .(6)
Substitute the present values into Equ.(6) and check if the resulting value of PEtop,bottom agrees with that obtained in Exercise 2.
Exercise 4. Substitute Expression (1) for the kinetic energy into Equ.(2) to obtain an expression for the potential energy at time t. Into the resulting equation for PE(t), substitute Expression (4) for E and simplify. You should obtain the following equation for PE(t):
PE(t) =
(m/2)A2w2 [1 - sin2
wt] .(7)
Using Equ.(5), this can be rewritten as
PE(t) =
(k/2)A2 [1 - sin2
wt] .(8)
Simplify Equ.(8) by using the standard trigonometric identity
sin2 q + cos2 q = 1 .(9)
Equ.(8) will simplify to
PE(t) =
(k/2)A2 cos2
wt .(10)
Check that the graph you sketched in Exercise 1 is consistent with this equation. Explain in words why you think the two are consistent. Discuss whether Equ.(10) has the right behavior at the top, bottom, and equilibrium positions of the motion.
Play and Pause the motion at some instant, and use Equ.(10) to calculate the value of PE at this instant. Compare your result to that shown in the Data box.
In Exercise 3 of the section "Kinetic Energy as a Function of Time", you were asked to write down a general expression for the position of the block as a function of time, assuming the motion starts at t = 0 with the block in its top position.
The equation is
y(t) = A cos
wt .(11)
Substitute this expression for y(t) into Equ.(10) to obtain an expression for the potential energy as function of position, instead of time. The result will be
PE(y) = (k/2)
y2 .(12)
Exercise 1. Sketch PE vs. y between y = -A and y = A. What is the name of this kind of a curve?
Exercise 2. Discuss if Equ.(12) is consistent with Equ.(6).
Exercise 3. RESET the applet. Set the spring constant to k = 200 N/m, and keep the mass and the amplitude at their default values of m = 1.25 kg and A = 0.20 m, respectively. These settings were used previously.
Drag the block all the way to the bottom, by clicking on the block and dragging. The block should be at y = -0.20 m. Calculate the corresponding value of PE from Equ.(12), and compare your result to the value in the Data box.
Drag the block upward to y = -0.15 m, and again calculate PE. Observe the blue potential energy column and the rate at which it changes. Continue doing this for y = -0.10 m, y = -0.05 m, y = 0, and then for positive y in steps of 0.05 m. Make a table of the values of PE vs. y. Discuss where the rate at which PE changes is greatest: near the bottom of the oscillation, or near the top or near the equilibrium point.
For each 0.05-m interval, calculate the average slope DPE / Dy of the potential energy curve over the interval from y to y + 0.05 m, and compare this average slope to the net force acting on the block at the midpoint of this interval, i.e., at y + 0.025 m. To calculate the net force, use the expression for the net force Fy derived in the Lesson accompanying the applet "Simple Harmonic Motion (Vertical Spring)",
Fy = - k y .
(13)
Answer. You should find that
Fy,mid = - DPE / Dy(14)
for each of the 0.05-m intervals. Note the minus sign in this equation. When the slope of the potential energy curve is positive, the net force is negative. Does that make sense?
You could make the y-interval smaller and smaller and would find Equ.(14) to be true always. In the limit of vanishing interval size, the average slope over an interval would become the slope at a point. This way one can prove that
The net force acting on the block at a given displacement y from equilibrium is equal to the negative slope at that point of the potential energy taken as a function of y.
The mathematical symbol for slope (also called derivative), as applied to the present case, is dPE / dy. With this notation,
Fy = - dPE /
dy(15)
Equ.(6) above gives the energy of the spring-block-earth system in terms of the amplitude A, and the spring constant k. Note that the energy depends on the square of the amplitude, not the first power of the amplitude.
Exercise 1. RESET the applet, and set the spring constant to k = 200 N/m and the amplitude to A = 0.10 m. Display the Data box.
Record the value of the energy E. Vary the amplitude with the slider from 0.10 m to 0.30 m, and record the corresponding values of E for several values of the amplitude along the way. Plot a graph of E vs. A and demonstrate that it exhibits the quadratic dependence on A from Equ.(6).
Comment. The quadratic dependence of the energy on the square of the amplitude is typical of all simple harmonic oscillations. E.g., electromagnetic radiation consists of waves that locally are harmonic oscillations. Thus, the intensity of electromagnetic radiation, i.e., the amount of energy transmitted per unit time, is proportional to the square of the amplitude of the oscillations of the electromagnetic field of the radiation.