# notes 1

```In PHYS 111/121 developed equation of motion for constant
acceleration
πβ π‘
πβ
(remember vector so acceleration constant in all direction)
π₯
π₯
π£ π‘
π£
π£
ππ‘
π£
π£
1
ππ‘
2
2π π₯
π₯
What happens if πβ not constant =&gt; the equations do not apply
Many real objects behave this way
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leaf blowing in the wind
wind up cars
teacher (me!)
pendulum (velocity keeps changing direction)
Let’s discuss a special case of nonβconstant acceleration: Simple Harmonic motion –
special kind of periodic motion
What is periodic motion
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motion that repeats itself at regular intervals
Example
o tapping finger to music
Periodic motion that moves back and forth over the same path is oscillatory motion – has 2 precise end
points
Example – piston in a car engine
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Good windshield wipers
Periodic motion is hard to get in realβlife friction can damp the motion
o brings end points close together
Called a damped oscillation – will not deal with this in class
“least complicated periodic motion”
Simple harmonic motion (SHM)
Simple harmonic oscillation (SHO)
Period T is the repeat time
π₯ π‘
π΄πππ  ωπ‘ describes SHM
A β amplitude of motion (m)
π (omega) angular frequency of motion (rad/sec)
π
2ππ
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f frequency of motion (cycle/sec or 1/sec or Hz (hertz))
How many times a second the motion repeats
T period of motion = how long it takes one full cycle to complete (sec)
π
1
π
So ωπ
ω
2π
π
2π
2ππ
What type of force gives SHM?
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Linear restoring force
Consider mass spring system on frictionless surface
Unstretched spring
Mass at equilibrium position
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x=0
F=0
Unstretched spring
Pull mass a distance to the right
Spring force pulls mass to the left towards
equilibrium
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x&gt;0
F&lt;0
Pull mass to the left
Spring force pushes mass to the right towards
equilibrium
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x&lt;0
F&gt;0
This is a restoring force – wants to restore position to equilibrium
Hooke’s law πΉ
ππ₯
negative sign means it’s a restoring force
Linear in position (not x or √x)
Today we want position as a function of time
i.e. what is the position at any given time?
2 ways of doing this
1. Cheating (like the book does with the circle) see appendix at the end
2. The true honest method
Consider mass spring system
Force not constant so can’t use equations of motion like last semester
πΉ
mπ
πΉ
mπ
ππ₯
π
⇒ N
ππ
mg
m 0 so N
mg
ππ
π
π₯
π
Get acceleration as a function of position
What have we done so far
Stated SHM goes as a cosine wave
Stated SHM requires linear restoring force really want to prove this
Non constant acceleration so need to try other method
Remember (from calculus)
If position is π₯ π‘
π₯
Velocity π£
Acceleration a
Can write from force equations (see above)
ππ
π
ππ₯
π
π
π
Define ω
π/π
π
π₯
π
π π₯
ππ‘
ω π₯
Differential equation which as solution
π₯ π‘
π΄ πππ  ωπ‘
θ
π
π
ω
ν π‘
ππ₯
ππ‘
π΄ω sin ωπ‘
π π‘
πν
ππ‘
π΄ω cos ωπ‘
θ
θ
A amplitude of motion
x cannot be greater than A or smaller than – A
ω is angular frequency of motion (how many radius per sec the motion goes)
ω
π
π
π
1 π
2π π
π
1
π
2 ππ
2π
π
π
θ phase constant depends on initial conditions (when the stopwatch starts)
0)
θ Do you pull the mass and release (θ
Do you give the mass a kick at equilibrium θ
Do you pull the mass to the left (βA) and release θ
For this class x π‘
Asin ωt or x π‘
What do these functions look like
π
Acos ωt will not consider different initial conditions
X = A max
ν
π΄
ππ
π₯
π΄ω
0 πππ’πππππππ’π
ν
π΄
0
π΄ωππ
0 π πππππ ππ πππ’πππππππ’π
π₯
π΄ πππ₯
ν
0
π
π΄ω πππ₯
π₯
0
ν
π
π΄ω πππ₯
0 π πππππ ππ‘ πππ’πππππππ’π
π₯
π΄ πππ₯
ν
0
π
π΄ω
(max, spring fully compressed)
π₯
0
ν
π΄ω (max)
π΄
0 π πππππ ππ‘ πππ’πππππππ’π
Drawing on the left: Unstretched spring
Middle drawing: Equilibrium with mass hanging at equilibrium
πΉ
mg
mg
kΔπΏ
0
k ΔπΏ
Drawing at the right, spring stretched some arbitrary amount
πΉ
ππ
πΉ
ππ
mg
k ΔπΏ
π₯
ma
mg
kΔπΏ
ππ₯
ma
π
π
π₯
π
Just like before
ω
π
π
Just new equilibrium position
Now consider energy
Kinetic energy is &frac12;ππ£
πΎπΈ
ππΈ
Potential energy stored in spring &frac12;ππ₯
If we had vertical mass spring also need to consider gravitational potential energy ππΈ
(using as equilibrium position the position of the mass hanging on spring with no motion)
Initial energy = final energy = total energy
KπΈ
&frac12;ππ£
PπΈ
KπΈ
&frac12;ππ₯
PπΈ
E
&frac12;ππ£
&frac12;ππ₯
πΈ
πππ₯
What is total energy?
When mass is at the end point of motion a distance A from equilibrium
So total energy πΈ
πΎπΈ
ππΈ
&frac12;ππ΄
&frac12;π 0
&frac12;ππ΄ and this is conserved
So π‘ππ‘ππ ππππππ¦
&frac12;ππ΄
&frac12;ππ£
πΎπΈ
&frac12;ππ₯
Textbook solves for ν
from forces get π
ππΈ
π΄ –π₯
π₯ and the other equations of motion
Example 1
A spring has an unstretched length of 20cm. It stretches 5cm after a 200g weight is placed hanging
vertically from the spring. The spring is then used in a horizontal springβmass oscillator with mass 300
g. The mass is pulled 3 cm from equilibrium and released. Calculate π, f, T, A. What equations
describe π π , π π , π π ? What is total energy of the system? What is the maximum velocity? What is
the displacement at π
ππ seconds?
Consider forces acting on 200 gm mass.
ππ₯
π
π π
π π
π₯
π π
0.2 ⋅ 9.8
0.05
0
39.2 π/π
Mass pulled 3 cm = 0.03 m and released, so furthest displacement is 0.03 m. So amplitude is 0.03 m.
A = 0.03 m
.
Angular frequency, π
.
Frequency π
1
π
ππππππ π
.
1
1.82
1.82π
0.55 π
Equations defining motion
π₯ π‘
π΄ cos ππ‘
0.03 cos 11.4π‘
π π‘
π΄π π ππ ππ‘
0.03 ⋅ 11.4π ππ11.4π‘
π π‘
π΄ω πππ  ω π‘
0.03 11.4
πππ  11.4π‘
0.342π ππ 11.4π‘
3.9 πππ  11.4π‘
total energy E = KE + PE (no gravity so ignore mgh)
1
ππ£
2
E
1
ππ₯
2
Easy to calculate if one of v or x is 0.
Say v =0, this occurs when mass is at an extreme end of the motion, ie x = A
So πΈ
π 0
&frac12; 3.92 0.03
0.0176 π½
Remember m, kg, s
ππ΄
Maximum velocity occurs when displacement = 0 (i.e. x = 0)
1
ππ£
2
πΈ
1
ππ₯
2
Get max when x = 0
ππ£
So πΈ
0.0176
⋅ 0.3 ν
0.342π/π
Solving gives ν
OR
From equations of motion
ωπ΄
ν
11.4 ⋅ 0.03
0.342 π/π
Displacement a t = 10 sec
π₯ π‘
0.03πππ  11.4π‘
0.03πππ  11.4 ⋅ 10
0.0186 π
If you get β0.012, remember deg vs rad.
Example 2
The motion of a 1 kg mass attached to a spring is described by x(t) = 0.3 cos(5t)
What is the total energy of the system?
Remember general form of equation of motion
π₯ π‘
π΄πππ ππ‘
so A = 0.3 m and π
2 ways to solve this this
πΈ
πΎπΈ
ππΈ
1
ππ₯
2
1
πν
2
Easy to calculate when ν max (x = 0)
E
ν
1
m ν
2
ωπ΄
So πΈ
π ωπ΄
1
⋅ 1 5 ⋅ 0.3
2
= 1.125 J
OR
If at xmax ( π
πΈ
0 ππ π₯
π΄
1
π π₯
2
1
ππ΄
2
ω
π
π π π
π
πΈ
ω ππ΄ (same equation as before
ω π
= 1.125 J
Example 3:
A bullet of mass 10 g travels at 600 m/s before striking (and embedding itself) into a 1 kg mass
attached to a horizontal spring with spring constant 25 N/m. What is the amplitude of the resulting
simple harmonic motion? Calculate frequency of oscillation. What is frequency if the bullet has mass 1
gm?
A little different as the simple harmonic motion starts with velocity at the equilibrium position, NOT by
pulling spring and releasing
From drawings 1 to 2: Conservation of momentum
Initial momentum = final momentum
π ν
π 0
0.01 600
so π£
π ν
π
0.01
.
1 π£
5.94π/π
.
total energy of the SHO (i.e. at 2) is
&frac12;ππ₯ We can calculate the energy just after the collision and then the
πΈ πΎπΈ ππΈ &frac12;ππ£
energy at the amplitude position to calculate A
Or
ωπ΄
We can use ν
ω
π
π
π£
π£
.
ω
2π
4.975 πππ/π ππ
5.94
.
π΄
f
25
1.01
4.976
2π
1.19π
0.79π ππ
If mass changes to 1 g, what happens?
Amplitude changes and frequency changes to (no need to calculate all the momentum stuff as the
frequency does not depend on the amplitude)
F=
=
.
= 0.795 secβ1
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