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# Vibration comparison Answered

I'm trying to compare two wildly different measurements, but I don't know enough about the physics of sound to even know what information I'm missing. I don't need exact numbers: some upper and lower limits would be fine, and perhaps even "back of the envelope" calculations in case the required information can't be found. I'd appreciate some advice on the formulas I need and what measurements am I still missing.

Object A is producing a sound whose intensity measures 25 dB at a distance of 1 meter.  This sound is, roughly speaking, a band-limited sawtooth wave with a fundamental frequency of about 35 Hz.

Object B is being driven up and down a distance of about 0.5 mm with an acceleration of 0.3 times that of gravity. The sound doing this is a sine wave with a frequency of 35 Hz.

I'd like to compare the peak accelerations experienced by the two objects.

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## 8 Replies

Josehf Murchison (author)2013-10-16

When you are trying to measure the forest, start by measuring one tree.

Comparing decibels to hertz is like comparing amps to volts where decibels are pressure and hertz is fluctuation and time.

On your speaker you have the movement but not the volume of air movement (cubic inches of air being moved) or the co-efficiency of air to transmit sound.

Then at the mike you need to do the same, in all probability at the mike you are receiving a sine wave and the mike turned it into a saw tooth or at the speaker it is receiving a sine wave and transmitting a saw tooth wave.

If you want the acceleration you already have one 0.3 x 9.80 = 2.94 meters a second if you did the math right.

You just need to measure distance and time, (a = d+t).

So from the numbers you gave me you have a movement of 35 x times a second a distance of 0.0005 meters (0.5mm) or 35 x 0.0005 = 0.0175.

That means you have a movement of 0.0175 meters a second in one direction and 0.0175 meters a second in the other for a total movement of 0.035 meters a second.

Now if you want to know the length of the sound wave, it is speed of sound divided by frequency.

343.2 meters a second divided by 35 Hz = 9.8057142857 meters, so the length of one wave is 9.8 meters. Now the fun part the air did not move 9.8 meters just the sound wave is that long.

See one tree at a time.

Joe

2013-10-16

I appreciate you breaking down the problem for me, but I'm having a bit of trouble following your explanation.

I should first clarify one point that may have been confusing. Situation A and situation B are not in any way connected (though perhaps pretending that they are would simplify the problem) and the forces acting in them do not have the same waveforms at all. I'm just trying to find out whether the peak accelerations that the objects experience are comparable.

The peak acceleration of object B is, as you say, 0.3gee or 3 m/s/s. Unless I missed something, that needs no more calculation.

Object A is producing a sound that is most definitely not derived from a sine wave, so I'm not sure your simplification will work. Maybe I need to somehow find the maximum of the second derivative of the waveform in A?

But just for the sake of keeping things simple, I guess we can pretend that the wave A is a sine. If I understand you right, we have:

d= ? m maximum distance covered by the surface of object A in one period
(you used the value of 0.5mm, but that was the distance object B moved in one period)
f= 35 Hz frequency
v= ? m/s average velocity of the surface of object A

The formula you gave is then v=2d*f.
The units check: meters/second = meters * (1/second)

Then:

c= 343.2m/s speed of sound
l= 9.8 m wavelength

The next formula you gave is l=c/f.
That checks too:  meters = meters/second / (1/second)

But I'm still confused as to how that velocity formula works. It seems to me that velocity would be at a maximum during the steepest part of the sine wave and zero at the peaks and troughs of the wave. (That's assuming a sine wave, which it's not.)

I also still don't understand how velocity or wavelength will help me find peak acceleration (m/s/s) for object A.

And what happened to the decibels measurement? I can convert that to RMS pressure, but I have no idea whether it's appropriate to do so. Dimensional analysis and algebra alone can't do physics.
p= 0.000020 kg/m/s/s * 10^(25 dB/20)
p= 0.00036 kg/m/s/s RMS pressure at 1m from object A

2013-10-16

When measuring two different things find the common ground like movement and start there.

When you used db I assumed it was sound you were working with.

Sound and electricity are very similar.

Joe

2013-10-16

I'm sorry, I don't understand what electricity has to do with this.

2013-10-17

Your question is about measuring two different things, and sound and electricity are two different things.

I am saying sound and electricity behave in similar manners.

The effect of sound can be perceived at 9.8 meters through air but the air didn’t move 9.8 meters.

The effect of electricity can be perceived at 9.8 meters through a wire but the wire didn’t move 9.8 meters.

When measuring two different things find the common ground (points of similarity) like movement or conductivity and start there

Joe

iceng (author)2013-10-17

Get an accelerometer kit from Parallax.com
And a SX USB stamp uP to run iT.

Software is free and easy.... sample programs are provided to log data peaks if you wish..

Then compare the two vibration with real data.

steveastrouk (author)2013-10-16

Without dimensions, materials/densities you don't have enough to start with.

2013-10-16

Yes, I know that I am missing significant pieces of information but don't know which. I can see that much by simple dimensional analysis.

That's why I would like some advice on choosing an appropriately simplified model and the formulas I will need. My question is about how to go about setting up the problem, not the final result.
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