1. Simple Harmonic Motion
Sinusoidal Curve and Circular Motion

2. A mass is oscillating on a spring
Position in
equal time intervals:
This type of motion, is called Simple Harmonic Motion. The system is trying to place the mass at its static equilibrium position

3. Repetitive Motion
A is the equivalent to x the displacement from equilibrium and is also the radius of the circle
Repetitive motion can be modelled as circular motion

4. What Can Algebra Tell Us?
Circle
Spring
This tells us that the period only depends upon the mass and the spring constant, not on the distance the spring is pulled back. In English, the time for the mass to move back and forth is ALWAYS the same even if the amplitude decreases.

5. What Can Algebra Tell Us?
Circle
Spring
This tells us that the maximum velocity depends upon the inverse of the mass, the spring constant, and the distance the spring is pulled back. In English, the maximumvelocity of the spring depends on how far it was pulled pack and the spring constant as well as the bigger the mass the slower the velocity of the spring.
We could has also come up with the equations by setting the Kinetic Energy equal to the Spring Potential Energy and solving for v

6. What Can Algebra Tell Us?
Circle
Spring
This tells us that the spring experiences no acceleration when r=0, this is, at the relaxed position. But we have to use logic to determine which way the acceleration is pointing.

7. Calculus Can Tell us More
Circle
From Force by Spring
The force by the spring is called a Restoring Force
But, the question is: What function’s second derivative is equal to this value?
Therefore x is a function of time, that when given a time value, t, I can tell you the position x, where the mass is.

8. Circular Motion
Yes, But what if I didn’t Know Calculus?

9. Vertical position versus time:
Period T
Period T

10. Sinusoidal motion
Displacement (cm)
Time (s)
Period T

11. Sine function: mathematically
y=sin(x)
y=cos(x) y 1
π/2 π
3π/2 2π
5π/2 3π
7π/2 4π
9π/2 5π x -1 2π

12. Sine function: employed for oscillationsx y
π/2 π
3π/2 2π
5π/2 3π
7π/2 4π
9π/2 5π -1 1
y=sin(x)
Displacement y (m) A
y= A sin(ωt)
T/2 T 2T
Time t (s) -A

13. Sine function: employed for oscillations
Amplitude A is the maximum
distance from equilibrium
Maximum displacement A
ωT = 2π
Initial condition y(t=0) or
Angular frequency in rad/s
Starting from equilibrium:
y=A sin(ωt)
Starting from A:
y=A cos(ωt)
Starting from -A:
y=-A cos(ωt)
Note:
Therefore

14. Position, Velocity, Acceleration
Phase Shift is determined by where the sinusoidal function starts.

15. Example 1 - determine y(t)
y(cm) 30
t(s) 5 10 15
Period?
T=4 s
Sine/cosine?
Sine
Amplitude?
15 cm
Where is the mass after 12 seconds?

16. Example 2 – graph y(t) Amplitude? 3cm y(t=0)? -3cm
When will the mass be at +3cm?
1s, 3s, 5s, …
Period? 2s
When will the mass be at 0?
0.5s, 1.5s, 2.5s, 3.5 s …
y (cm) 3 2 4 6 8
t(s) -3

17. Summary Harmonic oscillations are sinusoidal
Motion is repeated with a period T
Motion occurs between a positive and negative maximum value, named Amplitude
Position: y=A sin(ωt) or y=A cos(ωt)
Velocity: v=Aωcos(ωt) or v=-Aωsin(ωt)
Acceleration: a=-Aω2sin(ωt) or a=-Aω2cos(ωt)
Angular frequency:

18. Energy of Harmonic Motion
According to the law of conservation of energy, when the mass of attached to a spring is released, the total energy of the system remains constant. This energy is the sum of the elastic potential energy and the kinetic energy of the spring

19. Energy Example
A 55 g box is attached to a horizontal spring (force constant 24 N/m). The spring is then compressed to a position A= 8.6 cm to the left of the equilibrium position. The box is released and undergoes SHM.
What is the speed of the box when it is at x=5.1 cm from the equilibrium position?
What is the maximum speed of the box?
Can you solve these questions two ways?
Using energy?
Using position and velocity formulas?

20. Energy Example
A 55 g box is attached to a horizontal spring (force constant 24 N/m). The spring is then compressed to a position A= 8.6 cm to the left of the equilibrium position. The box is released and undergoes SHM.
What is the speed of the box when it is at x=5.1 cm from the equilibrium position?

21. Energy Example Let the position function be:
What is the speed of the box when it is at x=5.1 cm from the equilibrium position?
Let the position function be:
Let the velocity function be:
Set the position function = to 5.1 and solve for the t the time.
Plug this t value into the velocity function
Therefore the speed is 1.4 m/s

22. Energy Example
A 55 g box is attached to a horizontal spring (force constant 24 N/m). The spring is then compressed to a position A= 8.6 cm to the left of the equilibrium position. The box is released and undergoes SHM.
b) What is the maximum speed of the box?
The maximum speed will occur when it is at the equilibrium position (xf=0)

23. Energy Example b) What is the maximum speed of the box?
The maximum speed will occur when it is at the equilibrium position (xf=0)
Let the position function be:
Let the velocity function be:
Set the position function = to 0 and solve for the t the time.
Plug this t value into the velocity function
Therefore the speed is 1.8 m/s

24. Conservation of Energy
When the spring is in Equilibrium with gravity, the forces are balanced (the downward force of gravity matches the upward for of the spring. So:
Where ‘d’ is the equilibrium extension of the spring and ‘g’ is the acceleration due to gravity. Therefore : d

25. Conservation of Energy
To simplify things. I’m going to choose Ug = 0 at the extension equilibrium. I’m free to choose any location for this.
We will now set the spring bouncing. At any time the total energy is:
d= extension from relaxed d
x=0 +x
We pick a point below the equilibrium location so, mgx is negative.
Ug=0
This gives us:

26. Conservation of Energy
Now, let’s expand out this equation:

27. Conservation of Energy
Now, let’s expand out this equation:
I’m a constant
Therefore, I’m a constant too!

28. Java Site for SHM

29. What about Gravity in Vertical Harmonic Oscillations?
Let’s find the velocity at x. After we have pulled spring down to A.
In #2, we have a new equilibrium point where Spring force balances Gravity force.

30. What about Gravity in Vertical Harmonic Oscillations?

31. What about Gravity in Vertical Harmonic Oscillations?
Therefore on a vertical oscillator, we need not worry about gravity.