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Superposition

Why can you draw an equilibrium line anywhere and get the same answer?

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Original post by MMarìc
Why can you draw an equilibrium line anywhere and get the same answer?


Do you have an example to show what you mean? I am not sure what you mean by getting the same answer.
Reply 2
Original post by Eimmanuel
Do you have an example to show what you mean? I am not sure what you mean by getting the same answer.


Hi, it is for that one
Original post by MMarìc
Hi, it is for that one


The main point is because you are just shifting the curve up or down when the equilibrium line anywhere. It will not affect the way we add the displacements of the waves or the overall shape of the resultant wave.

I find that the question is a bit "misleading". I would get back to it when I have more time. Sorry about it.

You can try to model the second waves as follow
wave 1 : 2sin(x)
wave 2 : sin(2x)
Resultant : 2sin(x) + sin(2x)
Reply 4
Original post by Eimmanuel
The main point is because you are just shifting the curve up or down when the equilibrium line anywhere. It will not affect the way we add the displacements of the waves or the overall shape of the resultant wave.

I find that the question is a bit "misleading". I would get back to it when I have more time. Sorry about it.

You can try to model the second waves as follow
wave 1 : 2sin(x)
wave 2 : sin(2x)
Resultant : 2sin(x) + sin(2x)


Oh okay, my answer was about the principle of superposition. Thank you for explaining, and don’t worry about it :smile:
Original post by MMarìc
Oh okay, my answer was about the principle of superposition. Thank you for explaining, and don’t worry about it :smile:



I know you are learning about superposition of waves. But the question of “Explain why you can draw the equilibrium line anywhere and get the same answer.” is a bad question.

Equilibrium point in a wave profile has “real significance”. The “particles” at the equilibrium position have zero velocity. Equilibrium point allows us to define the amplitude of the wave.

When the equilibrium line is shifted, then the amplitude of the wave is changed. The particles on the new equilibrium line do not have zero velocity.

Have a look at the diagram below when the equilibrium line is shifted.

Resultant_wave_01.JPG


In this case, the equilibrium line is shifted down by 1 unit. The new resultant wave (say W2) has the same shape as the original resultant wave (say W1), the physics of W1 and W2 are different.

The particle at x = 0, for W1 has zero amplitude but W2 for has non-zero amplitude.

At the later part of your physics course or you may have learnt, intensity is directly proportional to the square of the amplitude of the wave.

Again, W1 and W2 will have different intensity at x = 0.

I would conclude that by drawing the equilibrium line anywhere and you will not get the same answer.

I think your teacher is asking “Explain why you can draw the “origin line” anywhere and get the same answer.”
Reply 6
Original post by Eimmanuel
I know you are learning about superposition of waves. But the question of “Explain why you can draw the equilibrium line anywhere and get the same answer.” is a bad question.

Equilibrium point in a wave profile has “real significance”. The “particles” at the equilibrium position have zero velocity. Equilibrium point allows us to define the amplitude of the wave.

When the equilibrium line is shifted, then the amplitude of the wave is changed. The particles on the new equilibrium line do not have zero velocity.

Have a look at the diagram below when the equilibrium line is shifted.

Resultant_wave_01.JPG


In this case, the equilibrium line is shifted down by 1 unit. The new resultant wave (say W2) has the same shape as the original resultant wave (say W1), the physics of W1 and W2 are different.

The particle at x = 0, for W1 has zero amplitude but W2 for has non-zero amplitude.

At the later part of your physics course or you may have learnt, intensity is directly proportional to the square of the amplitude of the wave.

Again, W1 and W2 will have different intensity at x = 0.

I would conclude that by drawing the equilibrium line anywhere and you will not get the same answer.

I think your teacher is asking “Explain why you can draw the “origin line” anywhere and get the same answer.”


Now that you explained it, I actually think now that she used a wrong word for the question and that it will not have a same answer if the equilibrium is moved. She haven’t handed back to us the worksheet yet so we haven’t talked about it that much since we had a lab earlier but maybe she will give it back to us tomorrow or on Friday. I’ll let you know her explanation :smile: . Thank you so much, I really appreciate your help!
Original post by MMarìc
Now that you explained it, I actually think now that she used a wrong word for the question and that it will not have a same answer if the equilibrium is moved. She haven’t handed back to us the worksheet yet so we haven’t talked about it that much since we had a lab earlier but maybe she will give it back to us tomorrow or on Friday. I’ll let you know her explanation :smile: . Thank you so much, I really appreciate your help!


Would be glad to know how your teacher answers this question. :smile:
Reply 8
Original post by Eimmanuel
Would be glad to know how your teacher answers this question. :smile:


Sorry to ask again, but do you know if the relationship between resonant lengths and wavelength change for different frequencies? I am guessing that it does change because if you change the frequency the resulting wavelength will change however i’m not sure about its relationship with resonant lengths.

And when we were doing the lab for Resonance Air Column we were hearing a louder sound at various lengths, is it because the frequency of the fork is getting the same frequency of water?
Original post by MMarìc
Sorry to ask again, but do you know if the relationship between resonant lengths and wavelength change for different frequencies? I am guessing that it does change because if you change the frequency the resulting wavelength will change however i’m not sure about its relationship with resonant lengths.

And when we were doing the lab for Resonance Air Column we were hearing a louder sound at various lengths, is it because the frequency of the fork is getting the same frequency of water?


Resonace_air_column01.png
Do you mean something as shown in the diagram above? If yes, let me know and I would elaborate it. Based on what you have said, you seem to get the ideas.
Reply 10
Original post by Eimmanuel
Resonace_air_column01.png
Do you mean something as shown in the diagram above? If yes, let me know and I would elaborate it. Based on what you have said, you seem to get the ideas.


Yes, that’s exactly it! :smile:
Reply 11
Original post by Eimmanuel
Resonace_air_column01.png
Do you mean something as shown in the diagram above? If yes, let me know and I would elaborate it. Based on what you have said, you seem to get the ideas.


Yes, it’s exactly like that! :smile:
Original post by MMarìc
Yes, that’s exactly it! :smile:



Original post by MMarìc
Sorry to ask again, but do you know if the relationship between resonant lengths and wavelength change for different frequencies? I am guessing that it does change because if you change the frequency the resulting wavelength will change however i’m not sure about its relationship with resonant lengths.


If you see the diagram above, say when

L1 = λ1/4

L2 = ¾ λ2

You use the same frequency tuning fork to produce the resonance, then λ1 = λ2, which means L1 L2.

Then L2 = 3L1.

If L1 = L2, you would need to use a different frequency tuning fork, then you would see that

λ1 = 2.

Original post by MMarìc
….And when we were doing the lab for Resonance Air Column we were hearing a louder sound at various lengths, is it because the frequency of the fork is getting the same frequency of water?



I am afraid that I don’t understand what you mean by “the frequency of the fork is getting the same frequency of water”. What do you mean by frequency of water?
Reply 13
Original post by Eimmanuel
I am afraid that I don’t understand what you mean by “the frequency of the fork is getting the same frequency of water”. What do you mean by frequency of water?


Actually, I am also confused about my answer. I am connecting it to Mechanical Resonance, the vibrating response of an object from a source that has the same frequency as the natural frequency of the object, but i just don’t know how and if they’re even related
Original post by MMarìc
Actually, I am also confused about my answer. I am connecting it to Mechanical Resonance, the vibrating response of an object from a source that has the same frequency as the natural frequency of the object, but i just don’t know how and if they’re even related


Here the system in response to the vibration of tuning fork is column of air not water. The water is to change the length of column of air.

There is one fundamental difference between the mechanical resonance (block–spring system) and the resonance in air column. (Teachers may not mention for all kinds of reasons)

In the mechanical resonance say the block–spring system, it has only one natural frequency but there are many resonance frequencies in the standing wave system of an air column.
Reply 15
Original post by Eimmanuel
Here the system in response to the vibration of tuning fork is column of air not water. The water is to change the length of column of air.

There is one fundamental difference between the mechanical resonance (block–spring system) and the resonance in air column. (Teachers may not mention for all kinds of reasons)

In the mechanical resonance say the block–spring system, it has only one natural frequency but there are many resonance frequencies in the standing wave system of an air column.


Okay. Then the mechanical resonance has nothing to do with the air column, but is it still because of the frequencies that makes a loud sound on every length?
Original post by MMarìc
Okay. Then the mechanical resonance has nothing to do with the air column, but is it still because of the frequencies that makes a loud sound on every length?



I am not sure what you mean “mechanical resonance has nothing to do with the air column”.

I assume you know the following experiment

https://www.youtube.com/watch?v=BSIw5SgUirg

The principles that explain the occurrence of standing wave or resonance is what you had learnt from mechanical resonance like block–spring system connected to a vibrating driver.

In the demo, when the frequency of the sine wave generator equals to one of the natural frequencies of the string, standing waves are produced and the string oscillates with a large amplitude.

In this resonance case, the wave generated by the sine wave generator is in phase with the reflected wave and the string absorbs energy from the sine wave generator.

If the string is driven at a frequency that is not one of its natural frequencies, the oscillations are of low amplitude and exhibit no stable pattern.

Hope it makes sense. :smile:
Reply 17
Original post by Eimmanuel
I am not sure what you mean “mechanical resonance has nothing to do with the air column”.

I assume you know the following experiment

https://www.youtube.com/watch?v=BSIw5SgUirg

The principles that explain the occurrence of standing wave or resonance is what you had learnt from mechanical resonance like block–spring system connected to a vibrating driver.

In the demo, when the frequency of the sine wave generator equals to one of the natural frequencies of the string, standing waves are produced and the string oscillates with a large amplitude.

In this resonance case, the wave generated by the sine wave generator is in phase with the reflected wave and the string absorbs energy from the sine wave generator.

If the string is driven at a frequency that is not one of its natural frequencies, the oscillations are of low amplitude and exhibit no stable pattern.

Hope it makes sense. :smile:


I apologize for the confusions, I am really bad at explaining and sometimes understanding some stuff. We have not talked about resonance that much yet, which causes my confusion.

Your explanations make sense. Again, thank you for your help!
Original post by MMarìc
I apologize for the confusions, I am really bad at explaining and sometimes understanding some stuff. We have not talked about resonance that much yet, which causes my confusion.

Your explanations make sense. Again, thank you for your help!


Don't have to be bad at not explaining thing well, I am not at explaining stuff too.

Resonance and superposition of wave at A level (I assume that your level of study now) are really difficult. Most of my students have the same feeling. :smile:
Original post by MMarìc
I apologize for the confusions, I am really bad at explaining and sometimes understanding some stuff. We have not talked about resonance that much yet, which causes my confusion.

Your explanations make sense. Again, thank you for your help!


I would recommend the physics textbook here.
https://openstax.org/subjects/science
You can download for free.
It usually good to read beyond your current school notes on physics.
I would sometimes make use of the conceptual questions in college physics to test my students understanding.

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