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Transcript
-Energy of SHM
-Comparing SHM
to Circular Motion
AP Physics C
Mrs. Coyle
Mrs. Coyle
Review: Gravitational Potential
Energy at large heights from the
Earth’s surface.
U = -GMm
r
The reference point
U=0 is at r=∞.
U <0 always.

An applied force does positive work
to bring the object to a position of
increased the potential energy.
Fg
The gravitational force, Fg, ,
always points in the direction
of lower potential energy.
Review: Gravitational Potential Energy at small
heights from the Earth’s Surface U=mgy.
We pick a reference level where U=0 when y=0.
If we pick a reference
level where U=0 at y≠0,
then we get a line with the
same slope but with a
different y intercept.
U
0
y
Fg
When an applied force does
positive work to move the
object, the gravitational force
does negative work.
An applied force does positive work
to bring the object to a position of
increased the potential energy.
The gravitational force, Fg, ,
always points in the direction of
lower potential energy.
Fs
Fa
Work Done
by an Applied Force Fa =kx
(opposing the spring force Fs=-kx)
to Increase the Potential Energy
of a Spring
𝒙
𝑭
𝟎 𝒂
Fs
W=
∙ dx= =
W= ½ k x2
𝒙
(𝒌𝒙)dx
𝟎
This work increased the potential
energy of the spring.
At the reference point x=0 , U=0
So at point x, U=½ k x2
Potential Energy of a Spring
An applied force does positive work
to bring the object to a position of
increased the potential energy.
Fs
Fs
The restoring force Fs (spring force)
always acts to bring the object to a
lower potential energy.
Remember: to find the spring
𝒅𝑼
force from U, Fs = 𝒅𝒙
When the spring is set to SHM,
the total mechanical energy of a SHM at any time is:
E= ½ mv2 + ½ kx2


From conservation of energy:
E= ½ mv2 + ½ kx2 = = ½ k A2 = constant
½ mv2 max = ½ k A2
SHM
Graph of
Potential and
Kinetic
Energy
vs Position
SHM
Graph of
Potential and
Kinetic Energy
vs
Time
Relating Circular Motion
and SHM
-Find v and a.
-Show that a= - w2 x (so is this SHM?)
-Find x max, v max, a max
x (t) = A cos (wt + f)
f
What type of motion with coordinates x and y
is described by the following two equations?

x (t) = A cos (wt + f)

y(t) = A sin (wt + f)
When the spring is set to SHM,
the total mechanical energy of a SHM at any time is:

E= ½ mv2 + ½ kx2
From conservation of energy:
E= ½ mv2 + ½ kx2 = = ½ k A2 = constant
½ mv2 max = ½ k A2

These can be used with vmax = w A =
𝒌
A
𝒎
to give:
½ mv2 max = ½ m w2 A2 = ½ k A2

At any given time, t
k 2
2
v
A

x


m
 w 2 A2  x 2
Ex # 23

A particle executes simple harmonic motion with an
amplitude of 3.00cm. At what position does its speed
equal half of its maximum speed?

Ans: Between 0.026m and -0.026m
Ex # 25




While riding behind a car travelling at
3m/s, you notice that one of the car’s
tires has a small hemispherical bump on
its rim.
A) Explain why the bump from your view
point behind the car, executes SHM
B) If the radii of the car’s tires are 0.300m,
what is the bump’s period of oscillation?
Ans: 0.628s