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I have used a new term spinning apparatus as I was unable to name it. I have tied a thread to a stone and was spinning it and I heard a sound something like that of a rotating propellor of a helicopter and then this idea stuck my mind

can i calculate it? if yes then how can any one suggest me something and also if any extra information any one needs regarding that apparatus feel free to comment

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If you frequency analyse the sound from your equipment then the fundamental frequency will the same as the rotational frequency. You could record the sound and use some software like Audacity to do the frequency analysis. This is exactly what alemi did in his reply to Can I compute the mass of a coin based on the sound of its fall?. Alternatively, and inevitably, there is an app for that.

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  • $\begingroup$ can you provide reference regarding how fundamental and rotational frequency are equal to each other $\endgroup$ – agha rehan abbas Jun 30 '14 at 12:25
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    $\begingroup$ @agharehanabbas - after one rotation the system is back in the same state as it was after the last rotation. While the motion of the system may be complex, you expect that there is no periodicity in the motion that is slower than one revolution. And that is (almost) the definition of "fundamental frequency". Try it - you will find that it works. $\endgroup$ – Floris Jun 30 '14 at 13:04
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    $\begingroup$ I'd have to experiment but I think the whole thing could be much more complicated. You also have to worry about what modes you're exciting on the motor casing! (though of course the driving one would dominate) Come to think of it, I'd love to see an animation of the frequency spectrum of an electric motor over various frequencies. $\endgroup$ – user12029 Jul 1 '14 at 0:29
  • $\begingroup$ @NeuroFuzzy go ahead bro $\endgroup$ – agha rehan abbas Jul 1 '14 at 13:23
  • $\begingroup$ @johnrennie would it be more complicated to analyse the tension in string or thread because by doing so we can actually lead to the calculation of mass of the body which is attached to the string $\endgroup$ – agha rehan abbas Jul 1 '14 at 13:26
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looking to the doppler effect: frequency equals (the speed of sound in what ever temperature air you are in: $c$)+(Velocity of you relative to the measured object: so if you are stationary this value is zero: $V(r)$)/(speed of sound in what ever temperature air you are in: $c$)+(Velocity of object relative to you $V(s)$: essentially the speed of the object you are measuring)=frequency: $f$

$f=(c+V(r))/(c+V(s))$; this rewrites to: $V(s)=(V(r)+c(1-f))/f$; if $V(r)=0$, then $V(s)=(c/f)-1$; resolving for f: $f=c/(V(s)+1)$

your $V(s)$ should equal: $(2πL_{rope})/(t_{2π})=V(s): (2πr)/t=V(s)$

Next comes the experimental set up. Your observer should be placed so that it forms a square with side lengths equaling the radius of your rope. This is so that when the object is coming directly towards the observer, and after π/2 interval of it's rotation is going directly away from you. With these two points of data we can extrapolate: $f=c/(c-V(s))$ when it is coming at you, and $f=c/(c-V(s))$ when going away from you. If this does not give the same V(s) after input of the observed frequency, then your object probably can't be solved for.

if you observe all the frequency over all time it should be semetric. Spendinding equal times in red/blue shift. Solve for the time when your object is at both points so you can know where to look in your data.

c in this: $c=331.4+.6(T°C)$

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  • $\begingroup$ Please consider typesetting your formulae in TeX, they are much easier to read that way. A basic tutorial is here. $\endgroup$ – ACuriousMind Jun 29 '14 at 20:28
  • $\begingroup$ what does the observer mean according to your answer $\endgroup$ – agha rehan abbas Jun 29 '14 at 20:30
  • $\begingroup$ that would be your detector. What ever instrument that is receiving the input. $\endgroup$ – Daniel Maksuta Jun 30 '14 at 23:42
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    $\begingroup$ @DanielMaksuta I think you misread the question. The question is not like an emitter of sound attached to a carousel, but like a motor in a black box turning an axle. $\endgroup$ – user12029 Jul 1 '14 at 0:24

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