1. Simple Harmonic Motion
Chapter 13**

2. Simple Harmonic Motion
An example of periodic motion in which the restoring force is linear.
A simple approximation that can be applied to many real world scenarios and allows for a straight forward analysis of these situations.
It can be applied to bridges, buildings, and many other real world objects.

3. Simple Harmonic Motion
Position oscillates following sinusoidal pattern
Amplitude (A) is the maximum distance that an object moves from its equilibrium position. A simple harmonic oscillator moves back and forth between the two positions of maximum displacement (A and –A).

5. FE = -kx (also called Fs)
Hooke’s Law Review
Elastic or Spring Force
FE = -kx (also called Fs)
k is the spring constant (N/m) that is unique to each spring/elastic
x is the displacement from the equilibrium position
**Note: The – part of the equation is because we are discussing the restoring force of the spring. The original force that caused the movement from equilibrium is the positive force (Newton’s 3rd Law: equal and opposite forces).

6. Stationary Mass on a Spring
Restoring force up:
Fres = Fe = -kx
In a vertical spring:
Fg = -Fres as the force pulling the spring down is gravitational force and the restoring force would therefore pull up.

7. Simple Harmonic Motion
Fe = Fe = -kx Fe = kx

8. Example 1
A piece of rubber is 45cm long when a weight of 8.0N hangs from it and is 58cm long when a weight of 12.5N hangs from it.
What is the spring constant of this piece of rubber?
*** Remember that x is the displacement from equilibrium position.

9. Solution

10. Check Your Learning
The spring in a typical Hooke’s Law apparatus has a spring constant of 15.0 N/m and a maximum extension of 10.0 cm. What is the largest mass that can be placed on the spring without damaging it?

11. Solution

12. SHM
Since restoring force is not constant, we cannot use kinematics equations
Period in SHM only depends on mass and spring constant (not amplitude or x)
Formula:
Where T = period (s)
m = mass (kg)
k = spring constant (N/m)

13. Example 2:
A vertical spring has a spring constant of 43.2 N/m. If a 135g mass is hung on it and made to vibrate,
a) what will the frequency of the vibration be?
b) what will its amplitude in this situation in comparison to when it had no fish on it?

14. Solution F= kx Fg = 43.2x mg = 43.2x (0.135)(9.81) = 43.2x
x = m or 3.06cm = Amplitude

15. Check Your Learning
When an 80.0kg person climbs into an 1100kg car, the car’s springs compress vertically by 1.2cm.
What causes the force on the spring?
What will be the frequency of vibration when the car hits a bump?

16. Solution

17. Practice
Worksheet: 1*, 2, 3, 4***, 5*, 6

18. Elastic Potential Energy
Work is done to stretch or compress a spring.
Remember this formula from grade 11:
Where E is energy (J)
k is spring constant (N/m)
x is the displacement from equilibrium position
***Remember that x can be replaced with Amplitude (A) as well!

19. More Grade 11 Review… Energy can be written as: Es (spring energy)
Ee (elastic energy)
Ep (potential energy)

20. Conservation of Energy
Energy is conserved. Energy is neither created nor destroyed.
Total Energy is made up of kinetic energy and the spring’s potential energy.

21. Conservation of Energy
Assuming there is no energy lost to friction (ie, we are only dealing with mechanical energy not Q),
Et = Ek + Ee

22. Things to Remember…
When the spring is pulled to its max (or compressed to its max compression) it has only potential energy (Ee).
Maximum acceleration occurs at the maximum compression or extension as it will change directions.
When released, it will have energy from both kinetic and potential.
At its equilibrium position, it will have only kinetic energy (so max velocity).
F = ma another equation that might help 

23. Example 3:
A spring with a spring constant of 125N/m is attached to a 245g mass. The mass is pulled a distance of 2.5cm from equilibrium and released.
What is the total energy of this system?
What is the maximum speed of the mass?

24. Solution

25. Check Your Learning
A block of mass 0.50kg is placed on a level, frictionless surface, in contact with a spring bumper (which has a 100.N/m spring constant). The spring bumper has been compressed by 0.30m. The spring, whose other end is fixed, is then released.
What is the speed of the block at the instant when the spring is still compressed by 0.10m?

26. Solution

27. Summary

28. Practice Worksheet: Skip 7
Hint for 10b: Remember that the bumper only compresses and goes back to equilibrium – it does not make a full period.
Hint for 11: Remember to use momentum here!