# How to visually determine the damping ratio when we know the peak amplitude

I am currently trying to understand this example from the book, where it visually calculates the dampening ratio given the peak amplitude as well as the natural frequency. Here is the graph from which a transfer function was obtained:

What they found out, is: "We estimate the natural frequency to be near the peak frequency, or approximately 5 rad/s. From the figure above, we see a peak of about 6.5 dB, which translates into a damping ratio of about 0.24"

How did they manage to get 0.24 by using the formula below? Is this a mistake in the book or am I not getting something?

• What did you calculate it to be? The formula is correct BTW. – Andy aka May 14 at 16:46
• damping noun,a reduction in the amplitude of an oscillation as a result of energy being drained from the system to overcome frictional or other resistive forces. dampen verb, (gerund or present participle: dampening) make slightly wet. I think you mean damping ratio rather than "dampening ration". – Transistor May 14 at 17:09
• @Andy aka Thank you for your comment. I used the formula but did not receive the correct answer. My mistake was that I forgot to convert the decibels using 10^( 6.5/20 ) – George Lua May 14 at 17:16
• @Transistor Did not even notice my mistake. I blame my frustration from this question... – George Lua May 14 at 17:17
• No problem. Even running a spill chick wouldn't have helped! – Transistor May 14 at 17:29

No, it's not a (major) mistake in the book. If I plug $$\\zeta\$$ = 0.24 in the formula, I get $$\M_P\$$ at 2.146 and, converting to decibels it is 6.63 dB. So not a million miles off 6.5 dB.