1. AP Physics
Section 11.1 to 11.3
Periodic Motion

2. Periodic motion
Motion that repeats in a regular cycle is known as periodic motion. Examples include a mass attached to a spring moving up and down, the motion of a pendulum (a weight on a string or rod), and a vibrating string such as the strings on a musical instrument.
There will be one point in the motion where the object is in equilibrium, and the net force is zero. When the object that is in motion moves away from this equilibrium position a restoring force will exist that will try to pull the object back to equilibrium.

3. Simple harmonic motion
If the restoring force, FS, is directly proportional to the displacement from equilibrium, the motion produced is called simple harmonic motion (SHM).
Two important properties describe simple harmonic motion, period and amplitude.
Period, T, measured in seconds, is the time needed to complete one full cycle. It is the seconds per cycle.
Amplitude, A, measured in meters, is the maximum displacement of the object from the equilibrium position.

4. Frequency Frequency units 1 f = T 1 Hz = 1/s
Frequency (f) — the number of cycles that occur in one second. Many physics texts use nu: ν.
Since period is seconds per wave, and frequency is waves per second, they are reciprocal values: 1
f = T
Frequency units
Frequency measures cycles per second, but “cycles” is a count, and is therefore unitless.
The standard units are therefore 1/s or s–1.
Another acceptable SI name for the unit is hertz (Hz).
1 Hz = 1/s

5. Mass on a spring
We saw the relationship between the force applied and the displacement in Section 6.4.
To review:
The displacement, x, is directly proportional to the applied force, FP.
FP ∝ x
A constant of proportionality lets us write the equation:
FP = kx
k is the stiffness constant.

6. As you push or pull on the spring, the spring supplies an equal reaction force, FS.
FS = –kx
Hooke’s Law
Recall that FS , tries to pull the mass back to equilibrium, and is called the restoring force.
This equation for an elastic object deformed from equilibrium is known as Hooke’s law. FS Fg

7. W W = Fx = (½kx)x W = ½kx2 PEelastic = US = ½kx2
As the equation and the previous image show, to create more displacement, more force must be supplied, and the restoring force gets larger as well.
Therefore, the force is a non-constant, but linearly increasing force. The work done will be the area under the F–x graph. The work equals the average force times the displacement.
F = ½(0 + kx)
W = Fx = (½kx)x
W = ½kx2
Therefore W
PEelastic = US = ½kx2
PE in the textbook
US on the AP test sheet

8. Mechanical energy of an oscillator
The total mechanical energy of the oscillator is:
Emechanical = K + US
Emechanical = ½mv2 + ½kx2
x is the position at any point in the motion, and v is the velocity at that point.
If the total energy is calculated from the maximum displacement, we can find the total energy in terms of amplitude, A. Velocity at A is zero, so:
Emechanical = ½kA2
The total mechanical energy of a simple harmonic oscillator is proportional to the square of the amplitude.

9. ½mv2 + ½kx2 = ½kA2 k k x2 v2 = (A2 – x2) = A2 1 – m m A2 k ½mvmax2 =
Therefore, we can write a single equation for mechanical energy:
½mv2 + ½kx2 = ½kA2
Solving this for v2, we get: k k x2
v2 =
(A2 – x2) = A2
1 – m m A2
At equilibrium, the system reaches its maximum speed, vmax. Therefore, K at this point equals E, since US = 0. k
½mvmax2 =
½kA2
vmax2
and = A2 m
Substituting this into the v2 equation gives: x2
v =
vmax
1 – A2

10. x0 vmax = = T = vmax k vmax A m
The last equation is exactly the same as the speed of a revolving object as seen from the side!
The maximum speed at x0 is the same as the speed of the object revolving in a circle.
The radius of the circle equals A. x0
circum.
2πA
vmax = =
period T
Solving for period:
2πA
T =
vmax
Substitute in the following value for vmax from the previous frame: k
vmax = A m

11. SHM in a weighted spring
So, the period of a mass spring system is given by:
T = 2π m k FS –A
Notice that period
does not depend
on A! It canceled. FS x0 a +A Fg Fg

12. Recall from section 8-1 that: ω = 2πf
The position, x, of the oscillator at any point is given by:
x = A cosθ x0 A θ θ x
Since
ω =
θ = ωt x t
x = A cos ωt
Recall from section 8-1 that: ω = 2πf
x = A cos 2πft
Since T is the reciprocal of f:
2πt
x = A cos T
Note: motion will repeat after t = T.

13. Graphing x as a function of t
The SHM function equation can be graphed for position as a function of time:
sinusoidal function

14. Sinusoidal functions x = A cos ωt 2πt x = A cos T x = A sin ωt 2πt
Note that we usually start the spring away from the equilibrium position at A or –A. This gives us the cosine equation:
x = A cos ωt
2πt
x = A cos T
Timing from equilibrium gives a sine equation. Both are sinusoidal.
x = A sin ωt
2πt
x = A sin 2T T