Simple Harmonic Equations

When solving for Simple Harmonic Equations, you will use the equations:

and

distance = d

amplitude = A

time = t

period =

frequency =

Putting the equations to use:

Ex 1) A weight attached to a spring is pulled down 6 inches from it's resting position and then released. If it takes 4 seconds for one oscillation, write the equation that relates the distance d from its rest position after time (in seconds). Assume no friction.

A = 6

t = 4

period = =

frequency = =

REAL Sinusoidal Functions

There are sinusoidal functions everywhere in the world. Here's an example on how the water tide shows sinusoidal similarities

Divine's Dock, South Carolina
33.5417° N, 79.0283° W

2010-02-18 5:05 AM EST 0.23 feet Low Tide
2010-02-18 6:57 AM EST Sunrise
2010-02-18 8:52 AM EST Moonrise
2010-02-18 10:22 AM EST 3.82 feet High Tide
2010-02-18 5:21 PM EST 0.10 feet Low Tide
2010-02-18 6:03 PM EST Sunset
2010-02-18 10:29 PM EST Moonset
2010-02-18 10:43 PM EST 4.19 feet High Tide
2010-02-19 5:47 AM EST 0.38 feet Low Tide
2010-02-19 6:56 AM EST Sunrise
2010-02-19 9:23 AM EST Moonrise
2010-02-19 10:57 AM EST 3.68 feet High Tide
2010-02-19 6:00 PM EST 0.16 feet Low Tide
2010-02-19 6:04 PM EST Sunset
2010-02-19 11:29 PM EST 4.19 feet High Tide
2010-02-19 11:29 PM EST Moonset

Like all sinusoidal functions, you find the equation first

1.) Sin or Cos? - cos

2.) Sinusoidal Axis -

3.) Amplitude - 4.19-.1=2.145

4.) Period - 12.12 hrs.

5.) Phase shift - 3.71

The equation

Precalculus 2/17/10

Finding Sinusoidal Functions by Hand

To write a sinusoidal functions by hand you must do four things:

1) Find the sinusoidal axis by finding the average of the highest and lowest data value

2) Find the amplitude by subtracting the highest data value by the sinusoidal axis

3)Determine the period using the formula

4)Determine any phase shift

In class we used the example of average tempertures per month from the city of Lyttelton, South Africa. The data values were: January-71, February-69, March-68, April-62, May-57, June-51, July-51, August-57, September-62, October-66, November-68, and December-69.

Finding the sinusoidal function:

1) Sinusoidal axis= 37.5

Highest data value is 64, and the lowest data value is 11. The average of 64 and 11 is 37.5

2) Amplitude= 26.5

The difference between the highest data value (64) and the sinusoidal axis (37.5) is 26.5

64-37.5= 26.5

3) Period is:

because the pattern repeats every 12 months and when we put 12 in as the variable B in the formula for the period, it reduces to this:

4) Phase shift= (x-7) because when the data values are put into a scatterplot with the Month as the x-axis adn Temperature as the y-axis, it is shaped like a cosine graph that has been shifted 7 to the right.

So once you have all of those pieces you can plug them into the equation of a sinusoidal function with A representin the amplitude, B representing the period, C representing the phase shift, and D representing the sinusoidal axis.

This is the equation of the sinusoidal function of the monthly teperatures of Lyttelton, South Africa:

This is the sinusoidal function graph of the temperature of Lyttelton, South Africa

Precalculus 2/11/2010

Sinusoidal Graphs

y=Asin((x-C))+D

A= amplitude B= period C= phase shift D= sinusoidal axis

Remember to use Mr. Corn's 5 step process. It makes writing equations easier!

1. Sine or Cosine
2. Sinusoidal Axis
3. Amplitude
4. Phase Shift
5. Period

1. Cosine

It is possible to have different outcomes but I like to find a sine or cosine graph that starts at a specific point on the graph. I chose cosine because it is at pi.


2. +3
The green dotted line.

3. 2
The amplitude is from the sinusoidal axis to the top/bottom of the graph. The difference of 3 to 5 and 3 to 1 is 2.

4. -

The distance from the origin to my starting point. It is negative because you would subtract pi to get to the origin.



5.
11pi is where the graph repeats from the starting point. Subtracting pi from this, 10pi. REMEMBER this is your period, BUT you have to divide 2pi by 10pi to place it in the equation.

Now put everything together...


y=2cos((x-))+3


And again!





1. Cosine

Starting point- inbetween 4 and 2, which is 3.

2. +2

3. 6

4. -3

Starting point at 3, subtract 3 to get to the origin.

5.

The graph repeats at 13. Subtracting 3 from this you get a period of 10. Divde 2pi by 10 to get your 'formula ready' answer of pi over 5 (reduced).

Final:

y=6cos((x-3))+2

Have a fun LONG Weekend!

and

Don't forget Mr. Corn's Website here

Pre-calculus 2/8/10

Inverse Trig Functions




Here you are finding the angle/radian
Ask yourself what radian you get if you take the sine of it


In this case, pi over 6 is the principle value, meaning that this is the first value on the unit circle that will give you the answer 1/2.
-here I showed two answers but there really are infinite answers, although Math XL will only ask for the principle value.

You may also come across a problem like this:



First find the radian that will give you root 3 over 2:




Next find the cosine of pi over 3 and 2pi over 3:








Today we also learned a little bit about the graph of inverse sin
(When doing this on your calculator, make sure it is in radian mode)

Put into your calculator



Here .3 represents a y-value and .305 represents an x-value

Try graphing sin(x) in Y1 and .3 in Y2

If you trace the first intersection after zero you will fins that it's coordinates are (.305, .3)
And if you trace the next intersection (when the graph is going down) the point is (2.837, .3), which is the same as pi-.305

Important inverse sin formulas:(calc meaning the number you get in you calculator after doing inverse sin of .3)
For where the graph is going up (the side where (.305, .3) is):


For the other side:


Important inverse cosine formulas:






Mr.Corn's web page here

Precalculus 1/03/10

Here's the base graph for SEC

every odd integer is the domain


is the range

is the period


here's the basic points on the graph.






Here's the basic graph for CSC







are the basic points




is the period













this is the graph SEC(2x)

here are its points


is the domain when n is an integer

is the period









this is the graph 2SEC(x)
it streches it AND moves up two

are the points





is the period




HELPFUL HINTS Basic graphs have the same pattern of nulls and integers.
2 infront of the x streches it a bit AND moves it up to two.