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Lecture

# L20_energy.pdf

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University of Toronto St. George

Physics

PHY354H1

Erich Poppitz

Winter

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Lecture 20: Energy and Oscillations
▯ relevant text sections: Morin 5.1-5.2
Conservation laws
▯ Conservation laws are a central part of modern physics, both for prob-
lem solving and for physical insights.
▯ It is not the case that you need a conservation law to solve a system,
but often a conservation law gives you a shortcut to solving a problem
that helps you avoid the use of F = ma or other even more complicated
methods.
Conservation of energy in one dimension
▯ Consider one dimensional motion under a force that is a function of x
ONLY (constant forces okay too). In that case we can use our sneaky
substitution trick from Chapter 3 to remove time from the problem:
We have
F(x) = ma = m dv = m dv dx = mv dv (1)
dt dx dt dx
▯ Keep in mind that the velocity v = v(x). Solving by separation of
variables:
mvdv = F(x)dx
Z v Z x
0 0
mvdv = F(x )dx
v0 Zxx
1 2 1 2 0 0
mv ▯ mv 0 = F(x )dx
2 2 x0
▯ Now we de▯ne a function V (x) such that
dV
F(x) = ▯ )
dx
1 we can ▯nd V(x) for any F(x) using:
Z
x 0 0
V (x) = ▯ F(x )dx + V 0 with V (x 0 = V 0 (2)
x0
I can then isolate for the integral of F in this equation and plug it into
my energy equation:
1 2 1 2
mv ▯ mv = V 0 V (x0
2 2
▯ If we put all the 0 stu▯ on one side and the other stu▯ on the other
side we get:
1 2 1 2
mv 0 V =0 mv + V (x):
2 2
▯ Since the left hand side is a constant, and the right hand side is true
for an arbitrary x, it must apply for ANY x. This means that the RHS
of the equation is always equal to a constant, it is conserved.
▯ We de▯ne the mechanical energy as:
E = 1 mv + V (x) = K(x _) + V (x):
2
where V (x) is the potential energy and K = mv is the kinetic energy.
2
▯ We found above that the mechanical energy is ALWAYS constant when
you have a F = F(x). Thus we call F(x) a conservative force, because
objects moving in F maintain constant energy. NOTE: You ONLY
de▯ne a potential energy for forces that are conservative.
▯ We can ▯nd an explicit solution for any energy conserving system: since
1 2
mv = E ▯ V (x)
2
(remember E is a constant) then
r
v(x) = ▯ 2 [E ▯ V (x)] = dx
m dt
2 ▯ This is a ▯rst order ODE, unlike Newton’s law, which is second order!
(Of course, this is because we already integrated once to get the energy
equation).
▯ Using separation of variables, we can ▯nd:
Z x Z x
dx 0 dx0
t ▯ 0 = ▯ 0 = ▯ q :
x0v(x ) x0 2 [E ▯ V (x )]
m
From this we can obtain t = t(x). Also, if the expression is not too
gruesome, we can invert it to ▯nd x(t).
Potential Energy of a Spring
▯ If F(x) = ▯kx; where x is the departure of the spring from equilibrium,
1 2
then using equation (2) we ▯nd V (x) = 2kx and the mechanical energy
is
1 2 1 2
E = mx_ + kx :
2 2
▯ If the spring force is the only force acting, then we have simple harmonic
motion with frequency ! = k=m (old news by now). Notice that this
is actually the ratio of the coe▯cients of the KE and PE terms. In fact,
we can write
1 2 2 2
E = m(x_ + ! x ):
2
▯ In general, if we can write the energy equation in the form
1 1
E = a_ + bq 2 (3)
2 2
where q is some variable related to position and E is some constant,
2
we know that the frequency will satisfy ! = b=a. If you want proof,
you just need to work with the F = ma equation and our method for
▯t
solving by trying solutions Ae and solving for ▯. All simple harmonic
oscillators (SHOs) have quadratic potentials.
3 Potential energy of a small angle pendulum
▯ Okay, based on what we learned from the spring, we should be able
to ▯nd the frequency of the pendulum by ▯nding the potential energy,
writing out the mechanical energy equation and taking the ratio of the
coe▯cients in front of the position and velocity terms.
▯ Lets write the position in terms of angle. For the kinetic energy, we
need to di▯erentiate the position:
x = l▯ ) x _ = l▯
So the kinetic energy is K = 1mv = m(l▯) . _2
2 2
▯ Setting the potential energy to zero when the pendulum is at ▯ = 0, we
get the potential energy by integrating the force (gravity in this case):
Z h
V = mgdx = mgh = mgl(1 ▯ cos▯)
0
For the small angle approximation, if we use a Taylor expansion of
2
cos▯ ▯ 1 ▯ ▯ =2 we get:
V (▯) ▯ 1mgl▯ :
2
So the mechanical energy is
E = 1ml ▯ + mgl(1 ▯ cos▯) ▯ 1ml ▯ + m

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