1. Harmonic Motion
AP Physics C

2. Another way to look at SHM
WHAT DOES THIS MEAN? THE SECOND DERIVATIVE OF A FUNCTION
THAT IS ADDED TO A CONSTANT TIMES ITSELF IS EQUAL TO ZERO.
What kind of function will ALWAYS do this?

3. A SINE FUNCTION!

4. Putting it all together: The bottom line

5. Energy of the SHM Oscillator
Assume a spring-mass system is moving on a frictionless surface (i.e. the total energy is constant)
The kinetic energy can be found by
K = ½ mv 2 = ½ mω2 A2 cos2 (ωt + φ)
K = ½ kA2 cos2 (ωt + φ) given: ω2 = k/m
The elastic potential energy can be found by
U = ½ kx 2 = ½ kA2 sin2 (ωt + φ)
The total energy is E = K + U
E= ½ kA2 cos2 (ωt + φ) + ½ kA2 sin2 (ωt + φ)
→ sin2 (ωt + φ) + cos2 (ωt + φ) = 1
E = ½ kA2

6. Energy of the SHM Oscillator
The total mechanical energy is proportional to the square of the amplitude
E = ½ KA2
Energy is continuously being transferred between potential energy stored in the spring and the kinetic energy of the block

7. Energy in SHM, summary

8. The Physical Pendulum
A physical pendulum is an oscillating body that rotates according to the location of its center of mass rather than a simple pendulum where all the mass is located at the end of a light string.
It is important to understand that “d” is the lever arm distance or the distance from the COM position to the point of rotation. It is also the same “d” in the Parallel Axes theorem.

9. Example
A spring is hanging from the ceiling. You know that if you elongate the spring by 3.0 meters, it will take 330 N of force to hold it at that position: The spring is then hung and a 5.0-kg mass is attached. The system is allowed to reach equilibrium; then displaced an additional 1.5 meters and released. Calculate the spring constant and angular frequency:
Spring Constant
110 N/m
Angular frequency
4.7 rad/s

10. Example
A spring is hanging from the ceiling. You know that if you elongate the spring by 3.0 meters, it will take 330 N of force to hold it at that position: The spring is then hung and a 5.0-kg mass is attached. The system is allowed to reach equilibrium; then displaced an additional 1.5 meters and released. Calculate the amplitude, frequency, and period:
Amplitude
Stated in the question as 1.5 m
Frequency and Period
0.75 Hz
1.34 s

11. Example
A spring is hanging from the ceiling. You know that if you elongate the spring by 3.0 meters, it will take 330 N of force to hold it at that position: The spring is then hung and a 5.0-kg mass is attached. The system is allowed to reach equilibrium; then displaced an additional 1.5 meters and released. Calculate the total energy and maximum velocity:
Total Energy J
Maximum velocity
7.05 m/s

12. Example
A spring is hanging from the ceiling. You know that if you elongate the spring by 3.0 meters, it will take 330 N of force to hold it at that position: The spring is then hung and a 5.0-kg mass is attached. The system is allowed to reach equilibrium; then displaced an additional 1.5 meters and released. Calculate the position of max velocity, maximum acceleration, position of max acceleration:
Position of mass at maximum velocity
At the equilibrium position
Maximum acceleration of the mass
m/s/s
Position of mass at maximum acceleration
At maximum amplitude, 1.5 m

13. Example – simple pendulum
A simple pendulum 4 m long swings with an amplitude of 0.20 m.
Compute the (i) period and (ii) frequency of that pendulum.
T = 2π*sqrt(L/g) = 4.01s; f=1/T = 0.249Hz
Compute the linear velocity of that pendulum at its lowest point.
Vmax=ωA = sqrt(g/L)A = 0.313m/s

14. Example – physical pendulum
A thin uniform rod of length L and mass M is pivoted about an axis through the rod at a distance of L/4 from one end.
Find the moment of inertia of the rod in terms of M and L.
I = Icm + md2 = mL2/12 + m(L/4)2 = 7mL2/48
Find the period of oscillation of this rod in terms of M, L, and fundamental constants.
T=2π*sqrt(I/(mgd)) = 2π*sqrt(7L/(12g))
How would the period be affected if the mass of the rod is doubled without changing its length? Explain.
No change
How would the period be affected if the length of the rod were doubled without changing its mass? Explain.
increase by sqrt(2)