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

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1. Mar 21, 2017

### DanRow93

1. The problem statement, all variables and given/known data
Determine the FIRST maximum and minimum values of the underdamped oscillation:

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

2. Relevant equations
3. The attempt at a solution

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!

2. Mar 21, 2017

### Staff: Mentor

Be careful here. Write an equation for x.

What is the periodicity of tan?

3. Mar 21, 2017

### DanRow93

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?

4. Mar 21, 2017

### Dick

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 $x$?

5. Mar 22, 2017

### DanRow93

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?

6. Mar 22, 2017

### Dick

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.

7. Mar 22, 2017

### DanRow93

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?

8. Mar 22, 2017

### Dick

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 $n=-1$??

9. Mar 22, 2017

### DanRow93

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.

10. Mar 22, 2017

### Dick

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.

11. Mar 23, 2017

### DanRow93

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