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- Thread starter SplinterCell
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Uploading...In summary, this conversation is about whether changing the frequency of the ultrasonic waves changes the phase difference between the waves. The answer is that it doesn't, but it does change the distance between the waves.

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Go to this grapher and play with s (proportional to frequency) and c (phase shift).

Hope this helps!

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Thank you. So what you're saying is that changing the frequency does not change the phase difference? But isn't increasing frequency (i.e. decreasing the wavelength) equivalent to increasing the distance from the emitter to the receiver (whichjamalkoiyess said:

Go to this grapher and play with s (proportional to frequency) and c (phase shift).

Hope this helps!

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It changes the distance between every peak and the other. Forget about the formulas for a bit and look at the graph if you can. Changing 's' which is the frequency is totally independent of 'c' the phase shift, which stays the same. In other words, the distance between peaks changes, but it changes in the same way for both and so the change between phases stays the same.SplinterCell said:But isn't increasing frequency (i.e. decreasing the wavelength) equivalent to increasing the distance from the emitter to the receiver (whichdoeschange the phase shift)?

Of course! but that has nothing to do with the speed of sound. A higher frequency holds more energy, and so for the same distance, it can host more peaks, which is an indirect proof that the speed is constant since everything is stable and the additional energy put into the wave by increasing the frequency goes into the number of peaks and not into the velocity.SplinterCell said:n both cases you have an increase in the amount of peaks/troughs that fit between the emitter and the receiver.

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@boneh3ad would you then say that my assessment is wrong? Just to be sure if what I said was right.

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Yes, but that's only true if we ignore the space component of the wave. The general equation is ##y(x,t) = A \cos (kx-\omega t+\varphi)##. In other words ##c = kx + \varphi##, but ##k## certainly does depend on the frequency. Also, I ask my question because I actually did such an experiment and the phase shift did in fact change with frequency. The only thing I fail to understand is how the speed of sound can be calculated knowing ##\Delta t## for different values of ##f##.jamalkoiyess said:It changes the distance between every peak and the other. Forget about the formulas for a bit and look at the graph if you can. Changing 's' which is the frequency is totally independent of 'c' the phase shift, which stays the same. In other words, the distance between peaks changes, but it changes in the same way for both and so the change between phases stays the same.

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Unfortunately the lab manual is not available. But the point is this: there's some emitter that generates sound waves of certain frequency ##f## (that you can control). And you also have a receiver (at some fixed distance ##d## from the emitter). Both devices are connected to an oscilloscope. You see two sinusoids on the graph (amplitude as function of time) - overall they look the same but they (usually) have some phase shift. The idea is this: let's look at two adjacent "similar" points ("similar" in the sense that they have the same amplitude) on these two graphs - for example, let's look at two adjacent peaks on these two graphs. Since the waves are shifted in time, these two peaks have a time difference (delay) ##\Delta t##. Now, let's start increasing ##f## as we watch these two points. Their "distance in time" (again, time is the horizontal axis here) will change. The question is how can I express ##f## in terms of ##\Delta t##.jamalkoiyess said:

I was able to derive the following formula ##\frac{d}{v} - \Delta t = \frac{1}{f}##, but it doesn't seem to give a good result of ##v## (having measured ##\Delta t## for different values of ##f## I tried to do a linear fit for ##1/f## as function of ##\Delta t## and look at the intercept), so I'm not sure about my derivation.

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What's the condition for ##\Delta t=0##?

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Indeed, in our experiment ##d \gg \lambda## (##d## is on the order of tens of centimeters whereas ##\lambda## is usually less than ##1 \text{cm}##).Ibix said:If I'm not mistaken that formula is only valid if ##d\leq\lambda##, which I suspect it's not

Well, ##\Delta t## is the time difference between the two peaks (which areIbix said:What's the condition for ##\Delta t=0##?

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That's a good question. We never had such situation, but I'd probably choose two adjacent peaks that are separated by the smallest non-zero distance. But I think the choice does not affect the overall dependence of ##\Delta t## on ##f##, am I wrong?Ibix said:

We measured ##\Delta t## in the following manner: choose one point (peak) on the "source sinusoid" and choose the closest point (peak) on the "received sinusoid". For example: here are the two sinusoids (similar to those that were displayed on the oscilloscope). So one way is choosing the "red" peak at ##t=0## and the "green" peak at ##t=-1.5##. As we change the frequency the horizontal distance between these peaks changes. That horizontal distance is the ##\Delta t##.

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You could also get an answer to the original problem all at once (rather than one frequency at a time) if you're clever. For example, by playing white noise through your speaker and calculating the cross-spectrual density between the two microphones.

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Unfortunately I'm not familiar with these concepts (and currently don't have too much time to learn them). I'm pretty sure there's an easier way to solve the problem.boneh3ad said:

You could also get an answer to the original problem all at once (rather than one frequency at a time) if you're clever. For example, by playing white noise through your speaker and calculating the cross-spectrual density between the two microphones.

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That's what I was trying to get at with the zero difference case. I was trying to get you to realize that your experimental protocol tells you that you should record ##\Delta t=0## in that case, but since ##d\neq 0##, ##\Delta t\neq 0## and there's something wrong since ##\Delta t## can't be both zero and not. But instead you decided to choose to do something different in that case. As a general rule, experiments that include "I make a decision" aren't measuring pure physics.

Think about it this way: if you adjust the frequency so that the received and transmitted signals are in phase, what is the relationship between ##d## and ##\lambda##?

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Right, but what I'm saying this is that we only need to care about the time shift between any two (initially) adjacent peaks. Here's how I see it: we can represent the "source wave" with the function ##g(t)=A\cos(-\omega t + \varphi_0)##, and the "receiver wave" with ##h(t)=A\cos (kd-\omega t + \varphi_0)##. Now, let's impose the "same amplitude" condition ##g(t_1)=h(t_2)##, which implies ##kx - \omega \Delta t = 2\pi n## where ##n## is some integer. Letting ##n=1## and dividing by ##\omega## we obtain ##d/v - \Delta t = 1/f##.Ibix said:The problem you have is that what you are measuring is not ##\Delta t##. Imagine turning your apparatus on. The source starts emitting a signal and a millisecond or so later the receiver picks it up. ##\Delta t## is that millisecond - right?

It must be some integer number of wavelengths. I understand that (and also mentioned it in one of my previous comments).Ibix said:Think about it this way: if you adjust the frequency so that the received and transmitted signals are in phase, what is the relationship between ##d## and ##\lambda##?

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But the point is that you have no way to determine n in your experiment. You don't know that it's 1, and in fact it isn't, which is why you're getting silly answers for the velocity. It's generally going to be of order ##d/\lambda##, I think.SplinterCell said:Letting ##n=1##

Right. Call the frequency ##f_0## and tune upwards until the next time the signals are in phase. Call this ##f_1##. How many wavelengths are there this time?SplinterCell said:It must be some integer number of wavelengths. I understand that (and also mentioned that in one the previous comments).

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I see my mistake now. The only way to know the correct value of ##n## would be to calculate ##d f / v##, but that would require knowing the velocity (which is what we're trying to find in the first place).Ibix said:But the point is that you have no way to determine n in your experiment. You don't know that it's 1, and in fact it isn't, which is why you're getting silly answers for the velocity. It's generally going to be of order ##d/\lambda##, I think.

##(n+1)\lambda##?Ibix said:Right. Call the frequency ##f_0## and tune upwards until the next time the signals are in phase. Call this ##f_1##. How many wavelengths are there this time?

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Slightly off topic: I remember walking beside a corrugated concrete embankment and noticing my shoe-steps and then my finger clicks were diffracted unto brief 'tweets', like lightning 'whistlers' but quicker. I've tried since, but found too much background noise to repeat the effect...

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More precisely, ##(n+1)\lambda_1##. And if you tune up to the next time they're in phase, it'll be ##(n+2)\lambda_2##. And what distance is each of these equal to?SplinterCell said:##(n+1)\lambda##?

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The same distance ##d##. Hence (substituting ##v=\lambda_i f_i##) we obtain ##v = d (f_2 - f_1) = d \Delta f##. Correct? So in general, if we manage to obtain a sequence of frequencies such that the phase shift for each one of them is zero (the waves overlap), we can calculate the velocity from the slope of ##f=\frac{v}{d} n## where ##n## is the index of the frequency in the sequence (of course we assume that we don't miss any overlaps in between). Thank you.Ibix said:More precisely, ##(n+1)\lambda_1##. And if you tune up to the next time they're in phase, it'll be ##(n+2)\lambda_2##. And what distance is each of these equal to?

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This is even easier if you learn the relatively easy concept of cross-correlation. If you have the right Matlab package there's even a built-in function (and it is standard in the academic version).

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