Download Simple Harmonic Motion ILAP

Simple Harmonic Motion ILAP
for Calculus I
(MATH 1401, Section 00x) [Modify your section #.]
Spring 2007
[Your name here.]
Introduction
“ILAP” stands for “Interdisciplinary Lively Application Project”. In this case, the
“Interdisciplinary” portion is taken from Physics. This project will focus on the motion
associated with a spring-mass system. We will show that we can accurately model this
motion which satisfies a first-order differential equation.
Spring-Mass Demonstration
We will collect position, velocity, and acceleration data from a spring-mass system (see
below, left). The sonic ranger and graphing calculator record the motion data (see below,
right).
We will analyze the total energy of the system. According to the Law of Conservation of
Energy, the total amount of energy (kinetic and potential) is the same at all times in this
system, ignoring air friction. We will show that a sinusoidal position function is the
solution to the associated differential equation.
[If you would like to add any comments about what you hope to achieve from writing up
this report, do so here. You could possibly receive 2 bonus points here.]
Kinetic Energy
This is the energy associated with a moving mass. The formula is
KE = (1/2)*m*v^2,
where KE is measured in [finish this part].
[Notice that I *centered* the formula by changing the justification, and then I left one
blank line and then changed the justification back to normal.]
In this system, the mass only moves vertically and y(t) measure the y-coordinate position
of the mass. Thus, we have
v = dy/dt.
[Explain what is happening when v is negative, and when v is positive. What does it
mean when the derivative is negative (or positive)? ]
Potential Energy in the Spring
As the spring stretches, potential energy (also measured in joules; all energy is measured
in joules) is stored in the spring. For a given mass, we find the equilibrium length of the
spring L measured in meters. When we measure the position of the mass, the equilibrium
position will be defined to y = 0.
This gives us an initial amount of potential energy in the spring. When the mass is lower
than this position [complete this part]
The change in the potential energy of the spring is given by the expression
(1/2)*k*(L – y)^2 – (1/2)*k*L^2.
If we expand and factor this expression, we obtain the more useful form
(1/2)*k*(???). [fill this part in]
The change in potential energy is measured in joules, k is a constant measured in
joules/meter^2. We will show how to calculate k later.
[Insert an example where y is negative. Is the change in potential energy positive or
negative? Do the same for when y is positive.]
Hooke’s Law specifically tells us how to find the potential energy stored in the spring. In
our demonstration, we showed that the amount of spring stretch is proportional to the
mass hung on the spring. If m is the mass, then the amount of stretch (x) is given by the
equation
9.8*m = k*x OR x = (9.8/k)*m.
[You were given the stretch values when m = 0.200 and 0.400 kg. Solve for k and give
its value here.]
[Here’s where you can earn more points. You may insert a page which shows a graph of
force vs. stretch. The graph is F = k*x. The area trapped beneath the force curve is the
energy necessary to stretch the spring. Thus, show that the accumulator function is
PE = (1/2)*k*x^2.
Explain why the formula (1/2)*k*(L – y)^2 – (1/2)*k*L^2 is actually a trapezoid!]
Gravitational Potential Energy
If we establish a certain height (near the surface of the Earth) where the potential energy
is defined to be zero, then the potential energy of a mass with respect to that height
depends on its vertical position. In this application, the height of our mass is [what?
How are we measuring it?].
The expression for the change in gravitational potential energy is
9.8*m*(change in y),
where the energy is measured in joules, m is measured in kilograms, and the change in y
is measured in meters. We note that the acceleration due to gravity near the Earth’s
surface is (-9.8) meters/second^2.
When the change in position is negative (the mass moves to a lower position), the system
loses gravitation energy, etc.
Conservation of Energy
The total energy of the system must be constant for all time (t is nonnegative). The total
energy for this system is
Total Energy = (1/2)*m*(dy/dt)^2 + (1/2)*k*(L - y)^2 – (1/2)*k*L^2 + m*g*y,
when y is measured relative to the “zero potential energy” height. The constant g is
9.8 m/sec^2.
We will show that the sinusoidal model will solve this equation if the total energy is
always a constant. Let
y = A*sin(t).
[Insert one page of engineering pad paper and show how to balance the equation.]
Thus, we must have k = [Type the correct expression here. This should agree with a
previous section item.]
From the print outs of the position vs. time graphs, we determined the value of P, the
fundamental period of the system when we hang a 0.2 kg mass on the spring. From our
derivation, we have also determined that the amplitude of the sinusoidal function does not
affect whether or not y = A*sin(t) is a solution to the differential equation.
Since
P = 2/,
we calculate the theoretical value of P using our derivation. That value of P is [fill it in
here] seconds. [How does this compare with the measured value?]
Conclusions
By balancing the different forms of energy associated with this spring-mass system, we
have determined the correct formula for calculating the period of any similar system,
knowing only the spring constant k, the mass of the oscillator, and the acceleration due to
gravity. More importantly, only a basic knowledge of Calculus I and College
Trigonometry was required.
[Explain why you can create a near-perfect clock with this mechanism! Make any other
interesting observations here and then wrap it up! BONUS POINTS!]