Answered: Use the Dahamel integral method to…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Question

Use the Dahamel integral method to derive expressions for the response of an undamped system subjected to the forcing functions shown in Figs. 4.46(a) to (c).

F(t)
F(t)
F(t)
F(1- cos
Fo
to ! 0
(a)
(b)
(c)
Fig.3

Expert Solution

Step 1

$x(t)=\frac{1}{m\omega }\int_{0}^{t}p(\tau )\sin \omega (t-\tau )d\tau$ (1)

Here x(t) is the response, p( $\tau$ ) is the force funtion, m is the mass and $\omega$ is the angular velocity. And $\tau$ specifically is the period of time over which which we have to fnd out the response.

For Figure 4.46(a) the force funtion is written as

F(t) = F₀ $0\leq t\leq t_{0}$ (2)

Here we will consider forcing system time as d $\tau$ not dt because we are determining response for a very small time and then we will integrate it over a whole time period to get a full response.

So equation (1) can be written as using the force function equation (2)...................

$x(t) = \frac{1}{m\omega }\int_{0}^{t_{0}}F_{0} \sin \omega (t-\tau )d\tau$ (3)

On integrating and substituting limits in the above equation (3) we get

$x(t) = -\frac{F_{0}}{m }[\cos \omega (t-t_{0}) - \cos \omega (t )]$ (4)

So the equation (4) reduced to

$x(t) = \frac{F_{0}}{m }[\cos \omega t-\cos \omega (t-t_{0})]$ Ans 4.46(a)

We have to solve the other two forcing functions in a similar way

For figure 4.46(b) the force function will be

$F(t) = \frac{F_{0}}{t_{0}}t$ (5)

So the Duhamel Integral expression can be written as

$x(t) = \frac{1}{m\omega }\int_{0}^{t_{0}}\frac{F_{0}}{t_{0}}\tau \sin \omega (t-\tau )d\tau$ (6)

on integrating the equation (6) and substituting the limits, we get

$x(t) = \frac{F_{0}\omega }{m\omega t_{0} }[-t_{0}\cos (t-t_{0})+\omega \sin (t-t_{0}-\omega \sin t)]$ (7)

Cancelling the commons, we have the response as

$x(t) = -\frac{F_{0}}{m t_{0} }[t_{0}\cos (t-t_{0})-\omega \sin (t-t_{0}+\omega \sin t)]$ Ans 4.46(b)

For figure 4.46(c) the force function will be

$F(t) = F_{0}(1-\cos \frac{\Pi t}{2t_{0}})$ (8)

So the Duhamel Integral expression can be written as

$x(t)=\frac{1}{m\omega }\int_{0}^{t_{0}}F_{0}(1-\cos \frac{\Pi \tau }{2t_{o}})\sin \omega (t-\tau )d\tau$ (9)

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