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Finding the time at which the displacement is a Maximum - Calculus

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Homework Statement



A damped oscillator has a displacement x as a function of time t given by x = e-t cos(∏t). Find the time at which the displacement is a maximum, when it is a minimum, and also nd
the time when x is a point of inflection.

Homework Equations



x = e-t cos(∏t)

The Attempt at a Solution



I've no problem differentiating the equation, its when I come to finding the value of T that I am having trouble with.

Please see the attachment.
 

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Answers and Replies

  • #2
arildno
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Well, you have identified that at t=infinity, there exists a stationary point. What is the displacement then?
A maximum or minimum?

You also have infinitely many other possibilities for stationary points, namely:
(pi)*sin(pi*t)+cos(pi*t)=0, meaning
tan(pi*t)=-1/pi

Which of these stationary points will represent maxima or minima?
 
  • #3
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When we use t=infinity the displacement (when substituting t=infinity in the equation) we get 0. The nature of the curve at this value is a 'Point of inflection' as shown in the attachment. Regarding the finding T for [∏sin(∏t) + cos(∏t) = 0] Arildno you put a minus sign in front of the ∏sin(∏t), shouldn't it be positive when looking through my workings out? When we do this we get Tan(∏t) = -1/∏.

Please do point out anything that seems incorrect so far...
-Thanks
 

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  • #4
arildno
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I hastily removed my minus sign, because it was, as you point out, incorrect.
 
  • #5
Ray Vickson
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Homework Statement



A damped oscillator has a displacement x as a function of time t given by x = e-t cos(∏t). Find the time at which the displacement is a maximum, when it is a minimum, and also nd
the time when x is a point of inflection.

Homework Equations



x = e-t cos(∏t)

The Attempt at a Solution



I've no problem differentiating the equation, its when I come to finding the value of T that I am having trouble with.

Please see the attachment.
Have you tried plotting a graph of x(t)? You should.
 
  • #6
arildno
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Note that the whole solution tends to 0 as t gets big.
Thus, your GLOBAL maximum should be your first LOCAL maximum, your GLOBAL minimum should be the first LOCAL minimum.
 
  • #7
Ray Vickson
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Note that the whole solution tends to 0 as t gets big.
Thus, your GLOBAL maximum should be your first LOCAL maximum, your GLOBAL minimum should be the first LOCAL minimum.
Your reasons are not sufficient in general, but they happen to lead to a correct result in this case. We can cook up functions that go to zero for large t, but whose maxima and/or minima occur at, say the 100th and 150th local max and min.
 

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