# The question is long, so I wrote it in the description part, and the you will find the name of the book. (Mechanical Engineering stuff)

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The question is long, so I wrote it in the description part, and the you will find the name of the book. (Mechanical Engineering stuff)
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I showed that connecting a pressure transducer having a natural frequency of 27,285 Hz (per Example 6.6 in Holman-now you know where the natural frequency of a pressure transducer comes from) to a ramjet combustor wasn’t going to pan out because the combustor has frequencies of 20 KHz, giving us a frequency ratio of 0.733, which, according to Fig 2.6 puts us dangerously close to resonance.  Now that is with an assumed damping ratio of 0.4, which is within reason.

Then I showed you that installing a ‘pigtail’ or a standoff tube would essentially eliminate the problem of ‘ringing,’ thus saving our prescious pressure transducer.  To illustrate that, I pointed you to equation 6.5 to calculate the natural frequency of the pigtail and to equation 6.6 to calculate the damping ratio.

Using the following calculate the natural frequency and damping ratio of the pigtail: tube diameter of 1/4″, tube length of 18″, pressure of two atmospheres, tube air temp of 200 C, and Example 6.2.  Convert to the required units as required and if I didn’t give you a parameter, use Example 6.2.

Calculate the frequency ratio between the pigtail and combustor, refer to Figure 2.6 using the damping ratio you have calculated and verify that you are out of resonance.  Make a copy of the figure and show me where you are by drawing a dot on the figure.
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Along the same lines as above, determine the natural freqency of a standard auto tire.  You will need to go the internet or library to find this out.  I prefer for you to find an equation much like 6.5 to describe the natural freqency of the tire.  This is going to be the forcing function frequency of the pressure transducer (the same one as in example 6.6 and use damping ratio of 0.4).  You will prove to yourself that this freqency is well below the natural freqency of the pressure transducer and that you are not ‘ringing’ the transducer.  Draw a dot on a copy of Fig 2.6 showing where you are.