Differentiation of damped motion function - Need help ly

[Shaybay92]

Differentiation of damped motion function - Need help urgently!

Homework Statement

Basically my task was to come up with a function to model the swing of a pendulum. The model I came up with was:

0.16e$$^{-0.25t}$$cos(($$\stackrel{2\pi}{1.22}$$)t-0.8) + 0.814​

The next part of my task asks me to find the point where the pendulum is first stationary. I attempted to differentiate this function, and set it equal to 0 to find t. When I graph it in my graphics calculator, I get approximately 0.18seconds, but when I do it by hand I get 0.165seconds. Help would be appreciated.

The attempt at a solution
When I differentiated using the product rule I got:

-0.04e$$^{-0.25t}$$ [cos((2$$\pi$$/1.22) t-0.8)-4(2$$\pi$$/1.22)sin((2$$\pi$$/1.22) t-0.8) ]

If you could please go through the word document attached, ths contains the working out of the derivative and solving when set to 0.

Help is greatly appreciated, thankyou!

 

[berkeman]

How can a pendulum have an offset in its displacement? Shouldn't it oscillate about 0 displacement? The initial phase in the cos() argument should be able to take care of an initial condition of an offset release...

 

[Shaybay92]

We made a model off experimental data. A motion detector measured the distance of the pendulum from it, and it did not exactly begin measuring when the pendulum was released either.

What do you mean the cos() argument?

 

[berkeman]

The argument to the cos() function in your equation has a time component and an offset constant phase component.

Did the pendulum oscillate about 0 displacement, or about 0.814m displacement? Zero is the bottom of the pendulum swing, right?

 

[Shaybay92]

The motion detector sat on a table 0.814m away from the centre of oscillation. The pendulum moved backwards and forwards, towards the motion detector.

 

[Shaybay92]

Do you understand what I mean? I'm not sure why I'm getting 0.15 off the actual value.

 

[Redbelly98]

Homework Statement

Basically my task was to come up with a function to model the swing of a pendulum. The model I came up with was:

0.16e$$^{-0.25t}$$cos(($$\stackrel{2\pi}{1.22}$$)t-0.8) + 0.814​

The next part of my task asks me to find the point where the pendulum is first stationary. I attempted to differentiate this function, and set it equal to 0 to find t. When I graph it in my graphics calculator, I get approximately 0.18seconds, but when I do it by hand I get 0.165seconds. Help would be appreciated.

0.165 is approximately 0.18. How accurate, in your estimation, is the value from the graph?

The attempt at a solution
When I differentiated using the product rule I got:

-0.04e$$^{-0.25t}$$ [cos((2$$\pi$$/1.22) t-0.8)-4(2$$\pi$$/1.22)sin((2$$\pi$$/1.22) t-0.8) ]

That looks right, and is zero at 0.165 s just as you calculated.

 

[Shaybay92]

The minimum was calculated using my graphics calculator. I graphed the function and pressed 'MIN' where it solved for the first minimum point at t=0.1805seconds. I realize it is close, however should it not be almost exact, as I kept all the values in the calculator without any rounding?

 

[Redbelly98]

When I differentiated using the product rule I got:

-0.04e$$^{-0.25t}$$ [cos((2$$\pi$$/1.22) t-0.8)-4(2$$\pi$$/1.22)sin((2$$\pi$$/1.22) t-0.8) ]

Wait, I just spotted an error. It should be a + in front of the sin term.

That changes things. Neither 0.165 or 0.18 are correct

p.s. I see your point about the calculator, I had thought you were just eyeballing the graph. My bad.

 

[Shaybay92]

It's negative because i factorized out -0.04? It was originally 0.016, so I had to make it -4.

 

[Redbelly98]

What is the derivative of cos?

 

[Shaybay92]

-sin?

 

[Shaybay92]

Oh! I think I see the problem ...lol

 

[Shaybay92]

I get a completely different answer... about 0.0048 or something... It made it worse :|

 

[Redbelly98]

I don't see anything squared here.

 

[Shaybay92]

Are you looking at the attached document?

0 =[cos(2π/1.22 t-0.8)+(8π/1.22)sin(2π/1.22 t-0.8) ]
That's after changing it back to a positive sin.. Then I let y = 2pi/1.22 t - 0.8, changed cos into a sin function using pythagorean identity, and then let w = sin y

-(8π/1.22)w= √(1-w^2 )
Then I squared both sides
[-(8π/1.22)w]= [√(1-w^2 )]^2

Which got rid of the +/- problem anyway

 

[Redbelly98]

Just saw your attachment. There's a much easier way to solve this.

If

0 = cosy + A siny​

then

cosy = ____ ? (solve above equation for cosy)​

and

tany = siny / cosy = ____ ?​

or

y = arctan(____?)​

 

[Shaybay92]

0 = cos y + (8pi/1.22)siny
sin y = -cosy/(8pi/1.22)
siny/cosy = -0.0485
tan y = -0.0485
y = -0.0485

Recall value of y:
(2pi/1.22)t - 0.8 = -0.0485
2pi/1.22t = 0.7514
t = 0.1459

This answer is still incorrect?

 

[Redbelly98]

t = 0.1459

This answer is still incorrect?

Nope, that answer looks good.

Not sure what happened with your calculator graph.

 

[Shaybay92]

How did you check that?

 

[Shaybay92]

I can't get that answer at all!

 

[Redbelly98]

How did you check that?

I entered expressions for both the function and the derivative in Excel, and looked for when (1) the function is a maximum and (2) the derivative is close to zero. I got t = 0.1459, to the nearest 0.0001.

 

[Shaybay92]

Is the cos function supposed to be cos((2pi/1.22)(t-0.8)) or (2pi/1.22)t-0.8 because this would change my calculations... if i leave it in brackets I actually get the second maximum (rather than first minimum) at 0.79sec

 

[Redbelly98]

I don't know, I'm just going by how you wrote it in post #1:

(2π/1.22)t - 0.8​

Looking at your data, about where does the first maximum appear to be?

 

[Shaybay92]

The value of 0.1459seconds is clearly wrong. In my table of comparisons with original data, 0.2seconds is a lower displacement than 0.1459seconds is. This can't be a minimum. It has to be 0.1805 but I just can't seem to find what I've done wrong in my differentiation. If anyone else can help it would be appreciated.

Thanks to the people who have already given their time to help me out.

 

[Redbelly98]

The 0.1459 s answer is the correct one for the equation you wrote, so if there is a problem it is not with your differentiation.

Perhaps that function was not a good fit to your data to begin with. You can try plotting your function from post #1 together with your data. Does it look like it's a reasonable function to describe your data? (It's a good idea to do this any time you are fitting a function to data.)