[Calc] First max and min values of an underdamped oscillation

[DanRow93]

Homework Statement

Determine the FIRST maximum and minimum values of the underdamped oscillation:

y=e^(-x/2)(4sin(3x)+3cos(3x)) cm

Homework Equations

3. The Attempt at a Solution [/B]
I firstly differentiated the above equation and got:

(-e^(-x/2)(22sin(3x)-21cos(3x)))/2

I checked this and it is correct.

Next I set this equation to = 0

I crossed off the exponential function (it can't equal 0?) and the /2 on the bottom.

22sin(3x)-21cos(3x)=0

21=22tan(3x)

tan(3x)=21/22

tan^(-1)(21/22)=0.7621

I plotted the graph online and checked this point, but it's definitely not a maximum or minimum, so I don't really understand; I thought that I could put dy/dx = 0 and that would tell me the points.

Also, this has only given me one point. How will I find the next point? Tan^(-1)(a number) only gives one definitive value.

I know that I can take the second derivative and that will tell me if the point is a maximum or minimum (+for minimum,-for maximum).

Can anyone help me understand how max and min works for this kind of equation? The oscillation gradually gets smaller, which I haven't ever done before.

Thank you!

 

[DrClaude]

tan(3x)=21/22

tan^(-1)(21/22)=0.7621

Be careful here. Write an equation for x.

Also, this has only given me one point. How will I find the next point? Tan^(-1)(a number) only gives one definitive value.

What is the periodicity of tan?

 

[DanRow93]

Be careful here. Write an equation for x.
What is the periodicity of tan?

Sorry that should be the value of 3x, so x=0.7621/3 = 0.254

And that does appear to be the first maximum!

Then I could prove that it is a maximum by taking the second derivative.

Tan has a period of Pi radians... Maybe I'm being stupid but I don't know how I find what the period is in terms of x.

When I do work it out though, I add half of one period to 0.254?

Edit:

tanx= tan(x+n*π)

So I put my value of x in here, but which number do I use for n?

I tried tan(0.254+(0.5*π)) but that wasn't right. n is an integer, but which integer?

 

[Dick]

Sorry that should be the value of 3x, so x=0.7621/3 = 0.254

And that does appear to be the first maximum!

Then I could prove that it is a maximum by taking the second derivative.

Tan has a period of Pi radians... Maybe I'm being stupid but I don't know how I find what the period is in terms of x.

When I do work it out though, I add half of one period to 0.254?

Edit:

tanx= tan(x+n*π)

So I put my value of x in here, but which number do I use for n?

I tried tan(0.254+(0.5*π)) but that wasn't right. n is an integer, but which integer?

Think about it. You've got $3x+n\pi=\arctan(\frac{21}{22})$. You've got the $n=0$ solution. What value of $n$ is going to give you the next largest value of $x$?

 

[DanRow93]

Think about it. You've got $3x+n\pi=\arctan(\frac{21}{22})$. You've got the $n=0$ solution. What value of $n$ is going to give you the next largest value of $x$?

Well l would think it would be n=0.5, seeing as 1π is one period of tan, and I am adding half of a wave length?

 

[Dick]

Well l would think it would be n=0.5, seeing as 1π is one period of tan, and I am adding half of a wave length?

You already said that doesn't work and it gives you a smaller value for $x$. $n$ needs to be an integer. Hint: $n$ can be negative.

 

[DanRow93]

You already said that doesn't work and it gives you a smaller value for $x$. $n$ needs to be an integer. Hint: $n$ can be negative.

I found the right answer. I did (3x+1π)/3

I'm still confused as to how this works though, because looking at the graph that I plotted of y=e^(-x/2)(4sin(3x)+3cos(3x)) it is only half of a wavelength from Maximum to Minimum, so why am I adding 1π, which is one whole wavelength of tan?

 

[Dick]

I found the right answer. I did (3x+1π)/3

I'm still confused as to how this works though, because looking at the graph that I plotted of y=e^(-x/2)(4sin(3x)+3cos(3x)) it is only half of a wavelength from Maximum to Minimum, so why am I adding 1π, which is one whole wavelength of tan?

You are confused because you appear to be simply guessing. Your starting point should be $3x+n\pi=\arctan(\frac{21}{22})$.
That's the equation that describes your extrema. You found the first one by setting $n=0$. If you set $n=1$ and solve for $x$ you will find that $x$ is negative, right? That's not it. How about trying $n=-1$??

 

[DanRow93]

You are confused because you appear to be simply guessing. Your starting point should be $3x+n\pi=\arctan(\frac{21}{22})$.
That's the equation that describes your extrema. You found the first one by setting $n=0$. If you set $n=1$ and solve for $x$ you will find that $x$ is negative, right? That's not it. How about trying $n=-1$??

I do understand the equation, so:

x = (arctan(21/22) + 1π)/3

I'm just stuck with how to know what n equals.

I know this is dumb because it's probably the simplest part of the whole task, thanks for being patient.

 

[Dick]

I do understand the equation, so:

x = (arctan(21/22) + 1π)/3

I'm just stuck with how to know what n equals.

I know this is dumb because it's probably the simplest part of the whole task, thanks for being patient.

That's ok. You may be looking for some deep answer. If you look at $3x+n\pi=\arctan(\frac{21}{22})$ you can see as $n$ increases, $x$ decreases. If $n$ decreases then $x$ increases. The smallest positive solution was at $n=0$. So the next positive solution will be at $n=-1$. The next one would be $n=-2$ etc etc. That's all there is to it.

 

[DanRow93]

That's ok. You may be looking for some deep answer. If you look at $3x+n\pi=\arctan(\frac{21}{22})$ you can see as $n$ increases, $x$ decreases. If $n$ decreases then $x$ increases. The smallest positive solution was at $n=0$. So the next positive solution will be at $n=-1$. The next one would be $n=-2$ etc etc. That's all there is to it.

Ok thank you! I guess I was looking into it too much.