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Amplification ratio for forced,damped motion

  1. Apr 8, 2009 #1
    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 in to 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 defintion gives
    [tex]
    \omega_0^2
    [/tex]
    in the numerator?
    Best regards
    John
     
  2. jcsd
  3. Apr 8, 2009 #2
    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.
     
  4. Apr 9, 2009 #3
    Hi Thaakisfox
    Do you mean that the correct definition is with the numerator = 1?
    Regards
    John
     
  5. Apr 9, 2009 #4
    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:
    Amplitude of steady state motion=\sqrt5:
    \omega_0^2=12
    [/tex]
    Then
    [tex]
    M=\frac{\sqrt5}{\frac{40}{12}}=\frac{3\sqrt5}{10}
    [/tex]
    I am confused.com!
    John
     
  6. Apr 9, 2009 #5
    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)
     
  7. Apr 9, 2009 #6
    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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