Amplification ratio for forced,damped motion

In summary: Hi again,I just looked at the source code for the function and it does indeed have the correct omega squared in the numerator.In summary, the book defines the amplification ratio, M, as:M=\frac{\omega_0^2}{\sqrt{(\omega_0^2-\omega^2)^2+(2r\omega)^2}}and an example is provided where F=40 and M=\frac{\sqrt5}{\frac{40}{12}}=\frac{3\sqrt5}{10}
  • #1
John 123
36
0
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 defintion gives
[tex]
\omega_0^2
[/tex]
in the numerator?
Best regards
John
 
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  • #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.
 
  • #3
Hi Thaakisfox
Do you mean that the correct definition is with the numerator = 1?
Regards
John
 
  • #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
 
  • #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)
 
  • #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
 

What is the amplification ratio for forced, damped motion?

The amplification ratio for forced, damped motion is a measure of the increase in amplitude of a system's response to a forcing function. It is typically represented by the symbol Q.

How is the amplification ratio calculated?

The amplification ratio is calculated by dividing the amplitude of the system's response by the amplitude of the forcing function.

What is the significance of the amplification ratio in forced, damped motion?

The amplification ratio is an important factor in understanding the behavior of a system under forced, damped motion. It can indicate the level of resonance and the potential for instability in the system.

How does the amplification ratio change with different damping ratios?

The amplification ratio is inversely proportional to the damping ratio. This means that as the damping ratio increases, the amplification ratio decreases, indicating a more stable response of the system to the forcing function.

Can the amplification ratio be greater than 1?

Yes, the amplification ratio can be greater than 1, indicating that the system's response to the forcing function is amplified. This is known as resonance and can lead to large amplitude oscillations and potential instability.

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