- #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
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