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:
<br /> M=\frac{\omega_0^2}{\sqrt{(\omega_0^2-\omega^2)^2+(2r\omega)^2}}<br />
I then solved the following problem in the book.
For what value of
<br /> \omega<br />
will the amplification ratio be a maximum? Find this maximum value?
MY ANSWER
I found the value of
<br /> \omega<br />
For which
<br /> \frac{dM}{d\omega}=0<br />
This value is
<br /> \omega=\sqrt{\omega_0^2-2r^2}<br />
which agrees with the book.
Then when you substitute this value into the formula for M you get
<br /> M_{max}=\frac{\omega_0^2}{2r\sqrt{\omega_0^2-r^2}}<br />
However the book answer gives:
<br /> M_{max}=\frac{1}{2r\sqrt{\omega_0^2-r^2}}<br />
The numerator has become 1 but their defintion gives
<br /> \omega_0^2<br />
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:
<br /> M=\frac{\omega_0^2}{\sqrt{(\omega_0^2-\omega^2)^2+(2r\omega)^2}}<br />
I then solved the following problem in the book.
For what value of
<br /> \omega<br />
will the amplification ratio be a maximum? Find this maximum value?
MY ANSWER
I found the value of
<br /> \omega<br />
For which
<br /> \frac{dM}{d\omega}=0<br />
This value is
<br /> \omega=\sqrt{\omega_0^2-2r^2}<br />
which agrees with the book.
Then when you substitute this value into the formula for M you get
<br /> M_{max}=\frac{\omega_0^2}{2r\sqrt{\omega_0^2-r^2}}<br />
However the book answer gives:
<br /> M_{max}=\frac{1}{2r\sqrt{\omega_0^2-r^2}}<br />
The numerator has become 1 but their defintion gives
<br /> \omega_0^2<br />
in the numerator?
Best regards
John