Amplification ratio for forced,damped motion

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Homework Help Overview

The discussion revolves around the amplification ratio, M, in the context of forced, damped motion as defined in a book on Ordinary Differential Equations. The original poster, John, is attempting to determine the value of ω that maximizes M and is comparing his findings with the book's answers.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • John attempts to find the maximum amplification ratio by calculating the derivative of M with respect to ω and setting it to zero. He questions the discrepancy between his result and the book's answer regarding the numerator of M_max.

Discussion Status

Participants are exploring the potential for typographical errors in the book's definition and answers. There is a shared sentiment that the definition of M may have inconsistencies, particularly regarding the numerator in the expression for M_max.

Contextual Notes

John references specific values and examples from the book, indicating that the discussion is grounded in the definitions and calculations presented in the text. The potential for errors in multiple editions of the book is noted, which may affect the understanding of the amplification ratio.

John 123
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Hi there
I am new to this forum but I am a regular contributor to the SOS maths forum.
I am working my way through a book on Ordinary Differential Equations and this book defines
the
Amplification Ratio,M, of the system under the above motion as:
[tex] M=\frac{\omega_0^2}{\sqrt{(\omega_0^2-\omega^2)^2+(2r\omega)^2}}[/tex]
I then solved the following problem in the book.
For what value of
[tex] \omega[/tex]
will the amplification ratio be a maximum? Find this maximum value?
MY ANSWER
I found the value of
[tex] \omega[/tex]
For which
[tex] \frac{dM}{d\omega}=0[/tex]
This value is
[tex] \omega=\sqrt{\omega_0^2-2r^2}[/tex]
which agrees with the book.
Then when you substitute this value into the formula for M you get
[tex] M_{max}=\frac{\omega_0^2}{2r\sqrt{\omega_0^2-r^2}}[/tex]
However the book answer gives:
[tex] M_{max}=\frac{1}{2r\sqrt{\omega_0^2-r^2}}[/tex]
The numerator has become 1 but their definition gives
[tex] \omega_0^2[/tex]
in the numerator?
Best regards
John
 
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It must be a typo mistake in the book, it happens quite a few times..
Even after 3-4 editions there might be mistakes in it.
 
Hi Thaakisfox
Do you mean that the correct definition is with the numerator = 1?
Regards
John
 
Hi again
The book quite categorically states:
The amplification ratio,M, as:
[tex] M=\frac{Amplitude of steady state output function}{\frac{Amplitude of input function}{\omega_0^2}}[/tex]
and so
[tex] M=\frac{\omega_0^2}{\sqrt{(\omega_0^2-\omega^2)+(2r\omega)^2}}[/tex]
Then an example is calculated where
[tex] F=40:<br /> Amplitude of steady state motion=\sqrt5:<br /> \omega_0^2=12[/tex]
Then
[tex] M=\frac{\sqrt5}{\frac{40}{12}}=\frac{3\sqrt5}{10}[/tex]
I am confused.com!
John
 
The thing is, he just writes it into the definition.

The answer for the M_max the book gave, is a typo, there should be \omega_0^2 in the numerator not 1. (you can also check the units, the units of the M differ from that of M_max)
 
Many Thanks Thakiisfox
I came to the same conclusion myself as I have seen the same formula in a book on Structural Engineering. Thus the typing errors must be in the answers[there are two answers that have 1 in the numerator and another two with the correct omega(0) squared in the numerator].
Regards
John
 

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