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

K.QMUL

## 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.

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.

#### Attachments

• 20131014_154740.jpg
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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?

K.QMUL
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

#### Attachments

Homework Helper
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I hastily removed my minus sign, because it was, as you point out, incorrect.

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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.

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.

Have you tried plotting a graph of x(t)? You should.

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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.