7.6 Modeling with trigonometric equations  (Page 4/14)

 Page 4 / 14

Simple harmonic motion

A type of motion described as simple harmonic motion    involves a restoring force but assumes that the motion will continue forever. Imagine a weighted object hanging on a spring, When that object is not disturbed, we say that the object is at rest, or in equilibrium. If the object is pulled down and then released, the force of the spring pulls the object back toward equilibrium and harmonic motion begins. The restoring force is directly proportional to the displacement of the object from its equilibrium point. When $\text{\hspace{0.17em}}t=0,d=0.$

Simple harmonic motion

We see that simple harmonic motion    equations are given in terms of displacement:

where $\text{\hspace{0.17em}}|a|\text{\hspace{0.17em}}$ is the amplitude, $\text{\hspace{0.17em}}\frac{2\pi }{\omega }\text{\hspace{0.17em}}$ is the period, and $\text{\hspace{0.17em}}\frac{\omega }{2\pi }\text{\hspace{0.17em}}$ is the frequency, or the number of cycles per unit of time.

Finding the displacement, period, and frequency, and graphing a function

For the given functions,

1. Find the maximum displacement of an object.
2. Find the period or the time required for one vibration.
3. Find the frequency.
4. Sketch the graph.
1. $y=5\text{\hspace{0.17em}}\mathrm{sin}\left(3t\right)$
2. $y=6\text{\hspace{0.17em}}\mathrm{cos}\left(\pi t\right)$
3. $y=5\text{\hspace{0.17em}}\mathrm{cos}\left(\frac{\pi }{2}t\right)$
1. $y=5\text{\hspace{0.17em}}\mathrm{sin}\left(3t\right)$
1. The maximum displacement is equal to the amplitude, $\text{\hspace{0.17em}}|a|,$ which is 5.
2. The period is $\text{\hspace{0.17em}}\frac{2\pi }{\omega }=\frac{2\pi }{3}.$
3. The frequency is given as $\text{\hspace{0.17em}}\frac{\omega }{2\pi }=\frac{3}{2\pi }.$
4. See [link] . The graph indicates the five key points.
2. $y=6\text{\hspace{0.17em}}\mathrm{cos}\left(\pi t\right)$
1. The maximum displacement is $\text{\hspace{0.17em}}6.$
2. The period is $\text{\hspace{0.17em}}\frac{2\pi }{\omega }=\frac{2\pi }{\pi }=2.$
3. The frequency is $\text{\hspace{0.17em}}\frac{\omega }{2\pi }=\frac{\pi }{2\pi }=\frac{1}{2}.$
3. $y=5\text{\hspace{0.17em}}\mathrm{cos}\left(\frac{\pi }{2}\right)\text{\hspace{0.17em}}t$
1. The maximum displacement is $\text{\hspace{0.17em}}5.$
2. The period is $\text{\hspace{0.17em}}\frac{2\pi }{\omega }=\frac{2\pi }{\frac{\pi }{2}}=4.$
3. The frequency is $\text{\hspace{0.17em}}\frac{1}{4}.$

Damped harmonic motion

In reality, a pendulum does not swing back and forth forever, nor does an object on a spring bounce up and down forever. Eventually, the pendulum stops swinging and the object stops bouncing and both return to equilibrium. Periodic motion in which an energy-dissipating force, or damping factor, acts is known as damped harmonic motion    . Friction is typically the damping factor.

In physics, various formulas are used to account for the damping factor on the moving object. Some of these are calculus-based formulas that involve derivatives. For our purposes, we will use formulas for basic damped harmonic motion models.

Damped harmonic motion

In damped harmonic motion    , the displacement of an oscillating object from its rest position at time $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ is given as

where $\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ is a damping factor, $\text{\hspace{0.17em}}|a|\text{\hspace{0.17em}}$ is the initial displacement and $\text{\hspace{0.17em}}\frac{2\pi }{\omega }\text{\hspace{0.17em}}$ is the period.

Modeling damped harmonic motion

Model the equations that fit the two scenarios and use a graphing utility to graph the functions: Two mass-spring systems exhibit damped harmonic motion at a frequency of $\text{\hspace{0.17em}}0.5\text{\hspace{0.17em}}$ cycles per second. Both have an initial displacement of 10 cm. The first has a damping factor of $\text{\hspace{0.17em}}0.5\text{\hspace{0.17em}}$ and the second has a damping factor of $\text{\hspace{0.17em}}0.1.$

At time $\text{\hspace{0.17em}}t=0,$ the displacement is the maximum of 10 cm, which calls for the cosine function. The cosine function will apply to both models.

We are given the frequency $\text{\hspace{0.17em}}f=\frac{\omega }{2\pi }\text{\hspace{0.17em}}$ of 0.5 cycles per second. Thus,

The first spring system has a damping factor of $\text{\hspace{0.17em}}c=0.5.\text{\hspace{0.17em}}$ Following the general model for damped harmonic motion, we have

$f\left(t\right)=10{e}^{-0.5t}\mathrm{cos}\left(\pi t\right)$

[link] models the motion of the first spring system.

The second spring system has a damping factor of $\text{\hspace{0.17em}}c=0.1\text{\hspace{0.17em}}$ and can be modeled as

$f\left(t\right)=10{e}^{-0.1t}\mathrm{cos}\left(\pi t\right)$

[link] models the motion of the second spring system.

difference between calculus and pre calculus?
give me an example of a problem so that I can practice answering
x³+y³+z³=42
Robert
dont forget the cube in each variable ;)
Robert
of she solves that, well ... then she has a lot of computational force under her command ....
Walter
what is a function?
I want to learn about the law of exponent
explain this
what is functions?
A mathematical relation such that every input has only one out.
Spiro
yes..it is a relationo of orders pairs of sets one or more input that leads to a exactly one output.
Mubita
Is a rule that assigns to each element X in a set A exactly one element, called F(x), in a set B.
RichieRich
If the plane intersects the cone (either above or below) horizontally, what figure will be created?
can you not take the square root of a negative number
No because a negative times a negative is a positive. No matter what you do you can never multiply the same number by itself and end with a negative
lurverkitten
Actually you can. you get what's called an Imaginary number denoted by i which is represented on the complex plane. The reply above would be correct if we were still confined to the "real" number line.
Liam
Suppose P= {-3,1,3} Q={-3,-2-1} and R= {-2,2,3}.what is the intersection
can I get some pretty basic questions
In what way does set notation relate to function notation
Ama
is precalculus needed to take caculus
It depends on what you already know. Just test yourself with some precalculus questions. If you find them easy, you're good to go.
Spiro
the solution doesn't seem right for this problem
what is the domain of f(x)=x-4/x^2-2x-15 then
x is different from -5&3
Seid
All real x except 5 and - 3
Spiro
***youtu.be/ESxOXfh2Poc
Loree
how to prroved cos⁴x-sin⁴x= cos²x-sin²x are equal
Don't think that you can.
Elliott
By using some imaginary no.
Tanmay
how do you provided cos⁴x-sin⁴x = cos²x-sin²x are equal
What are the question marks for?
Elliott