Amplification ratio for forced,damped motion

John 123
Messages
32
Reaction score
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:
<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
 
Physics news on Phys.org
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:
<br /> M=\frac{Amplitude of steady state output function}{\frac{Amplitude of input function}{\omega_0^2}}<br />
and so
<br /> M=\frac{\omega_0^2}{\sqrt{(\omega_0^2-\omega^2)+(2r\omega)^2}}<br />
Then an example is calculated where
<br /> F=40:<br /> Amplitude of steady state motion=\sqrt5:<br /> \omega_0^2=12<br />
Then
<br /> M=\frac{\sqrt5}{\frac{40}{12}}=\frac{3\sqrt5}{10}<br />
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
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top