PES 1110 Fall 2013, Spendier
Lecture 39/Page 1
Today:
- HW 9 due Friday
- Last HW 10 assigned, due next week Wednesday Dec 11th
- Last Quiz 6 Monday 12/9/2013 on HW 9! Quiz cover sheet online
- Finish Kepler's Laws (13.7)
- Oscillations (15.1-15.3)
Kepler’s Laws:
1) Each planet’s orbit traces out the shape of an ellipse with the sun located at one focus.
2) The imaginary line from the sun to a planet sweeps out equal areas in equal times.
- gave us conservation of angular momentum for a satellite orbiting the sun
So for any satellite
v p rp = va ra The planet goes fastest at Perihelion, since here the r is smallest.
3) The period of the planet’s motion is proportional to the orbit’s semi-major axis to the
3/2 power.
Kepler’s Third Law
3) The period of the planet’s motion is proportional to the orbit’s semi-major axis to the
3/2 power.
Kepler: T ∝ a 3/ 2 (derived from data)
Above that same power, r3/2, was in the period for the circular satellite.
Newton’s proof (I won’t show it here)
2πa 3/ 2
T=
GM sun
This is the same equation as for the period for circular motion with a (semi-major axis)
replaced with r (radius).
This result is interesting. There is nothing at the center of the ellipse. The sun is not at the
center. So we use this distance a where nothing resides but empty space to obtain the
period of a satellite orbiting in an ellipse.
PES 1110 Fall 2013, Spendier
Lecture 39/Page 2
Example 1:
A satellite is in elliptical orbit with a period of 8.00 x 104 s about a planet of mass 7.00 x
1024 kg. At aphelion, at radius 4.5 x 107 m, the satellite’s angular speed is 7.16 x 10-5
rad/s. What is its angular speed at perihelion?
Periodic Motion
Today we start chapter 15 and talk about periodic motion. Periodic motion is the
generalization of circular motion. It is also called an oscillation, so it is any sort of
repeated motion. If you go around a circle you repeat your motion each time you
completed one full period. A mass hanging from a spring also repeats itself when you let
it jiggle up and down. A pendulum swinging back and forth is repeated motion. The
pumping of your heart is also periodic.
DEMO: Mass on a spring
When I hang the mass from a spring, it stretches the spring a little bit and this position is
the equilibrium position for oscillation. The equilibrium position is where it is at rest. Be
careful, this is different from what we used before; the equilibrium position was given by
the length of the un-stretched spring when we calculated the force on a block and
potential energy of the spring.
PES 1110 Fall 2013, Spendier
Lecture 39/Page 3
Periodic Motion or Oscillation - Any repeated motion.
So a lot of terms we used in the past apply here as well and get generalized:
Cycle - One complete round trip (this is instead of one revolution on a circle)
Amplitude, A - Maximum displacement from zero.
Period, T - Time for one cycle. (like satellite motion)
Frequency, f - The number of cycles per time. (how frequently does the motion repeat
itself)
1
1
f =
Unit: =Hz (Hertz) (cycles is like radians it is unitless)
T
s
Simple Harmonic Motion
Simple Harmonic Motion (SHM) is a type of periodic motion you will see most
frequently in your studies. It is the simplest type of periodic motion and occurs for
example when a mass is connected to a spring with no friction.
(Same idea if vertical)
pull mass by a distance x to the right
Fel = kx
(elastic force of a spring, increases with stretching distance)
∑ F = ma (Newton’s 2nd law)
−Fel = max
(spring pulls to the left, only force since no friction)
−kx = max (here we started from un-stretched, x=0, Fel=0)
k
ax = − x
m
We know the relationship between instantaneous acceleration and position
dvx d 2 x
ax =
= 2
dt
dt
Hence
k
d 2x
= −  x this is the (second order) differential equation for SHM
2
 m 
dt
We need to solve this equation in order to figure out what the position is as a function of
time for a mass connected to a spring. Most second order differential equations are hard
to solve. But his one is pretty easy. You know the solution to this equation.
In calculus: f '' = −cf
In this case x, the position, is the function, we want to solve for. The second derivative of
some function is equal to minus a constant times the original function. Here time is t is
independent, so the function we are looking for is x(t). What function if you take the
derivative twice gives you back the same function with a negative sign in front of it?
PES 1110 Fall 2013, Spendier
Lecture 39/Page 4
Reminder:
d
sin(t ) = cos(t )
dt
d
cos(t ) = − sin(t )
dt
Hence the general solution for the differential
2
 k 
d 2x

 x
= −
dt 2
 m 
can be a combination of sine and cosine
 k 
 k 
x(t ) = A cos 
t  + B sin 
t 
 m 
 m 
But your book uses only the cosine solution by adding something called the “phase
constant or phase angle ϕ”. We will describe why we will need a phase angle shortly.
 k

x(t ) = A cos 
t + φ = A cos (ωt + φ )

 m
k
2π
=
= 2π f
m
T
A = amplitude of the motion
Angular frequency ω =
NOTE: c = ω 2 =
k
m
NOTE: nothing to do with angular velocity!
Aside: cos (α + β ) = cos (α ) cos (β ) − sin (α ) sin (β )
Hence: C cos (ωt + φ ) = C cos (ωt ) cos (φ ) − C sin (ωt ) sin (φ ) = A cos (ωt ) + B sin (ωt )
Proof:
PES 1110 Fall 2013, Spendier
Lecture 39/Page 5
Amplitude: A
Maximum distance from zero
Largest value of cosine is 1 and smallest -1
To modify the cosine function so it goes between the maximum magnitudes of the
displacement specific for the problem we multiply cosine by the amplitude A.
Phase Angle: ϕ
Units: rad.
We use the phase angle to shift the Cosine back and forward so that it starts wherever
needed.
A cos(t) has to start at t=0, with amplitude A
What if the spring did not start at its maximum value of the amplitude? Maybe we started
at 2/3 of its maximum value or another value. Hence we need to take this basic cosine
function and shift it to the left by ϕ (ωt+ϕ) or to the right by ϕ (ωt-ϕ) so x(t) starts where I
need it to start.
PES 1110 Fall 2013, Spendier
Lecture 39/Page 6
Angular Frequency: ω
I just gave you the equation
2π
ω=
= 2π f
T
1
frequency: f =
T
But why is this true?
It is important that we can modify the angular frequency of the cosine function since it
might take longer or shorter than T = 2π (if x=Acos(t)) for the motion to repeat itself.
PES 1110 Fall 2013, Spendier
Lecture 39/Page 7
General solution: x(t ) = A cos (ωt + φ )
A…. make sure it has the correct max amplitude
k
2π
=
- change m or k, or both
ω…. make sure we get the right period - ω =
m
T
ϕ…..make sure it starts at the right place in the cycle
Example 2:
A mass of 1.5 kg on a massless spring is experiencing simple harmonic motion on a
frictionless floor. It's maximum displacement from the equilibrium position is 12 cm ans
one period of oscillation takes 2.0 s.
a) What is the equation for this motion? Assume that at t=0, the mass has its maximum
negative displacement.
b) What is the spring constant?