Thread: sinusoidal equation

Any ideas of how to do a sinusoidal equation? My teacher didnt really teach this because she thinks that the book explains alot but it would help if shed explain it. So if anyone could explain this that would be soo great. steps and all so i can follow because thats how i learn

Originally Posted by takkun0486

Any ideas of how to do a sinusoidal equation? My teacher didnt really teach this because she thinks that the book explains alot but it would help if shed explain it. So if anyone could explain this that would be soo great. steps and all so i can follow because thats how i learn

A sinusoidal equation can be a sine equation or a cosine equation. The same sinusoidal graph can represent a sine curve or a cosine curve. [A cosine curve is a sine curve that is shifted horizontally by 90 degrees or pi/2 radians to the right. Or a sine curve is a cosine curve that is shifted horizontally by 90 degrees or pi/2 radians to the left.]

The basic sine curve starts at zero, or is zero at the origin (0,0) then it goes up to the maximum. The basic cosine curve starts at the maximum at the origin (0,0) then it goes down to zero. So, since the given exercise mentions the highest point and the lowest point, then we use the cosine curve for the problem. [The weight is not exactly at the highest point at t=0, yes, but since the highest point is mentioned first, I normally associate that with a cosine curve/equation.]

The basic cosine equation is in the form
y = cosX ------------(1)
where the ampiltude is 1; there is no horizontal shift or phase angle; no vertical shift.

The standard cosine curve with all the whistles is in the form
y = A*cos[B(x-C)] +D ------------(2)
where
--> A = amplitude---the maximum heigth or distance of the curve from the neutral horizomtal axis. Or, it is half of the vertical distance between the maximum and minimum points of the curve.
--> B = frenquency of the cycles or periods----number of cycles or periods per unit time
--> C = horizontal shift or phase angle----the angle where the basic origin is shifted horizontally; a value of "X" here where the basic origin is moved horizontally.
--> D = vertical shift----a value of "y" here where the basic origin is moved vertically.

----------------
Given: high point = 60cm above the floor; low point = 40cm above the floor.
So, amplitude A = (60 -40)/2 = 10cm ----**

Given: high point at 0.3sec; next low point at 1.8sec.
In a sinusoidal curve, from high point to next low point is half of a cycle or half of a period.
So, period = (1.8 -0.3)*2 = 3 sec/cycle.
Period = 360/B in degrees; Period = 2pi/B in radians.
Say, we use radians, hence
3 = 2pi/B
B = 2pi/3 -------**

Given: atfer 0.3sec, a high point.
That means the high point is "shifted" horizontally to the right by 0.3sec. [If no horizontal shift, the high point should be at t=0sec.]
So, C = 0.3 sec -----**

Again, given: high point = 60cm above the floor; low point = 40cm above the floor.
That means the basic neutral horizontal axis is halfway vertically from the 60-cm high point and the 40-cm lowpoint, which is (60 +40)/2 = 50cm above the floor.
That means further that the basic origin is vertically shifted 50cm above the ground.
So, D = 50cm -------**

Therefore, the cosine equation is
h = A*cos[B(t -C)] +D --------------(i)
where
h = height of weight above the floor at ant time t seconds, in centimeters.
t = any time, in seconds
A, B, C, D = as explained above in Eq.(2).

h = 10*cos[(2pi/3)(t -0.3)] +50 --------------answer.

----------------------------------------------------------
b) the distance of the weight from the floor at t = 17.2 sec.

h = 10*cos[(2pi/3)(17.2 -0.3)] +50
h = 58.15 cm --------------------------answer.

----------------------------------------------------------
c) the first 3 times when the weight is 53cm above floor.

53 = 10*cos[(2pi/3)(t -0.3) +50
53 -50 = 10*cos[(2pi/3)(t -0.3)]
cos[(2pi/3)(t -0.3)] = 3/10 = 0.3
angle [(2pi/3)(t -0.3)] = arccos(0.3)
(2pi/3)(t -0.3) = 1.266103673
t -0.3 = 1.266103673 / (2pi/3)
t -0.3 = 0.60452
t = 0.60452 +0.3
t = 0.90452 sec --------------the first time.
Since cosine is periodic, and the period of this particular cosine curve is found at 3sec/cycle, then every 3sec after 0.90452sec, the weight is 53cm above the floor, so,
0.90452 +3 = 3.90452 sec --------------the second time.
3.90452 +3 = 6.90452 sec -------------the 3rd time.

Originally Posted by ticbol

c) the first 3 times when the weight is 53cm above floor.

53 = 10*cos[(2pi/3)(t -0.3) +50
53 -50 = 10*cos[(2pi/3)(t -0.3)]
cos[(2pi/3)(t -0.3)] = 3/10 = 0.3 --------------positive cosine***
angle [(2pi/3)(t -0.3)] = arccos(0.3)
(2pi/3)(t -0.3) = 1.266103673
t -0.3 = 1.266103673 / (2pi/3)
t -0.3 = 0.60452
t = 0.60452 +0.3
t = 0.90452 sec --------------the first time.
Since cosine is periodic, and the period of this particular cosine curve is found at 3sec/cycle, then every 3sec after 0.90452sec, the weight is 53cm above the floor, so,
0.90452 +3 = 3.90452 sec --------------the second time.
3.90452 +3 = 6.90452 sec -------------the 3rd time.

Let me correct that.

The 3.90452 sec is actually the 3rd time, and the 6.90452 sec is the 5th time. The second time corresponds to the angle (2pi/3)(t -0.3) in the 4th quadrant.

Since cos[(2pi/3)(t -0.3)] is positive 0.3, then angle [(2pi/3)(t -0.3)] can be in the 1st or 4th quadrants.

In the 1st quadrant, the angle is 1.266103673 radians.
In the 4th quadrant, the angle is (2pi -1.266103673), which is 5.017081634 radians.
So,
angle [(2pi/3)(t -0.3)] = arccos(0.3)
(2pi/3)(t -0.3) = 5.017081634 ----------in the 4th quadrant
t -0.3 = 5.017081634 / (2pi/3)
t -0.3 = 2.395480026
t = 2.395480026 +0.3
t = 2.695480026 sec --------------the 2nd time.