How Does String Tension Affect the Sound of a New Musical Instrument?

[Emily_20]

Homework Statement

A new musical instrument is fashioned from a metal can of length L and diameter L/10, open at one end, with a string stretched across the diameter of the open end. The string is tensioned such that the 3^rd harmonic frequency of the vibrating string matches the fundamental frequency of the can.

a) What is the mathematical relationship between the speed v_t of transverse waves on the string and the speed v_a of longitudinal waves in the can?b) What happens to the sound produced by the instrument if the tension in the string is doubled? Describe what happens to both the pitch and the intensity, and defend your answer with an argument as quantitative.

Homework Equations

F_n = nv/4L, v=f *lambda

The Attempt at a Solution

a) 3v/4L=1v/4L
F_can= F_string

b) The frequency goes up.

I need help with these two questions.

 

[Simon Bridge]

a) 3v/4L=1v/4L

... You have just written that 3=1 (multiply both sides by 4L/v) - this is false.

The frequency formula you have written above does not work for both can and string.
You need to distinguish between the speed values for the can and the string.

If you sketch the wave-form, instead of relying on picking the right equation, you should be able to relate the lengths in question to the wavelengths of the waves.
The equation you want is $v=f\lambda$

 

[Emily_20]

The problem is that I do not even know how to sketch the wave-form. I do not even think that I understand what's the point of this problem.. maybe that's why I cannot solve it...

 

[Emily_20]

Here is an attempt for a sketch but I am not sure if its right..

 

[Simon Bridge]

Good - you got the wave-form on the string - now you can write out the wavelength in terms of L.
Notice that one wavelength fits 2/3rds of the string...

On the same diagram, you can sketch in the pressure wave for the can.

The point of the problem is to exercise your understanding. You get a lot of these puzzle-type things before you can be turned loose on a real example.

 

[Emily_20]

Hello sir, thank you for your help I appreciate it.
I wrote out the wavelength in terms of L for both can and string and I tried to write the mathematical relationship using it, is that correct?

 

[Emily_20]

I think I made a mistake in the calculation in V_t/L = 3/2 V_a/ L/15

 

[BvU]

In your original equation you had a 4. Now you have L = $\lambda_c$.
What is the wavelength that matches the fundamental frequency of the can ?

 

[Emily_20]

I'm not even sure if my original equation is correct. I though at the beginning that the string will act with a frequency for a tube with one open end and one closed end that's why I had a 4. For the wavelength of the can I used L=λ_c

 

[BvU]

Well, to me the description and your drawing look a lot like "a tube with one open end and one closed end" !

 

[Emily_20]

Thanks for the help I appreciate it. If I understand its right so the can is actually going to have a frequency of Fn = nv/4L and the string is going to have a frequency of f=V/λ

 

[BvU]

Do distinguish $\lambda_{string} = {2\over 3} \; {L\over 10}$ and $\lambda_{can} =$ ...
And: do distinguish $v_{string}$ and $v_{can}$.
What is equal is the tone, i.e. the ...

 

[Emily_20]

4L? because the can has a frequency of f₁ and the string is going to have a frequency of f₃

 

[Emily_20]

Simon Bridge said that I can sketch in the pressure wave for the can what does he mean by pressure wave?

 

[Simon Bridge]

If you prefer, sketch the logitudinal wave along the can.
The wave in the can is a sound wave in air... i.e. a wave of variations in air pressure.
You have notes that show the harmonics in a tube that is open at one end.

You can reason it out ... in the logitudinal wave, the air particles move the most at antinodes, and the least at nodes, where do those have to be? You are also told that the wave is the fundamental.

This way of reasoning through the physics is more reliable than trying to remember the right equation.

 

[Emily_20]

I got that the wavelength of the string is 2/3 (L/10) and the wavelength of the can is 4L is that correct?

 

[Simon Bridge]

Some notes... when you wrote. $$\lambda_s =\frac{2}{3}L=\frac{2}{3} \frac{L}{10}$$... that was nonsense. Can you see why?

You keep making that kind of mistake... it will cost you marks.
It is very important that you figure it out.

 

[Emily_20]

Unfortunately, I am very confused. Physics is not my strength. This problem is very important for me

 

[Simon Bridge]

I got that the wavelength of the string is 2/3 (L/10) and the wavelength of the can is 4L is that correct?

Can you explain how those should be correct? Reasoning is more important than being right.
The point of getting you to sketch things out is so you won't need to ask if its correct, there will only be one possibility.

 

[Emily_20]

I thought the the wavelength of the string is the diameter of the can. If the string has a 3rd harmonic frequency and its an open pipe I though using the equation λs=2/3L. For L I used the diameter of the can because its the length.

 

[Simon Bridge]

Unfortunately, I am very confused. Physics is not my strength. This problem is very important for me

The mistake you keep making is actually maths, not physics. Take another look, what is wrong with the equation?

 

[Emily_20]

λs=2/3L=2/3 L/10 Its going to be 2L/30 which is L/15

 

[Simon Bridge]

I thought the the wavelength of the string is the diameter of the can. If the string has a 3rd harmonic frequency and its an open pipe I though using the equation λs=2/3L. For L I used the diameter of the can because its the length.

... however L is defined in the problem to be the length of the can, not it's diameter.
In fact, you used it for both things in your maths. This is not allowed, different things have to have different symbols.

 

[Emily_20]

Ok so the equation is going to be λs=2/3L_s

 

[Simon Bridge]

... where Ls is the length of the string... and Ls = L/10. See how that works?

Personally Id have used D for diameter but only because I hate subscripts.

 

[Emily_20]

Is L/10 considered as the wavelength of the string too?

 

[Simon Bridge]

No. You seem to be having trouble understanding what equations are saying.
Ls says "the length of the string"
L says "the length of the can"

Ls = L/10 says "the length of the string is the length of the can divided by 10"

You already have an equation that says "the wavelength on the string is two-thirds the length of the string".

 

[Emily_20]

So I understand that the wavelength of the string is length of the can (L) divided by 15 is that right?

I got 15 from plugging L/10 for Ls in the equation λs=2/3Ls

 

[Simon Bridge]

Well done.
Basically you ended up with the same answer, but, this time, you arrived using clear maths.
How you get there is more important than the destination.

Now you need to relate what you know about the wavelengths to the va and vt values you are asked to find.

 

[Emily_20]

Ok Thanks for the help! I appreciate it

 

[Simon Bridge]

No worries, just be careful when you write down an equation... make sure it says what you mean.
Its OK to invent new variable names. You don't have to, and often shouldnt, use the variables in the "standard form" of a physics equation.
Also... get used to drawing pictures of everything. You'll notice your teachers do that all the time.

 

[BvU]

I got that the wavelength of the string is 2/3 (L/10) and the wavelength of the can is 4L is that correct?

Yes, it's correct -- in a way. But the fact that you have to ask worries me.

The wavelength of the standing wave in the string is 2/3 (L/10). L/10 is the length of the string. L is the length of the can. You'll do well not to use L otherwise -- it'll only cause confusion. See for example your post #20. And earlier on, Simon's reaction in post #17 to your post #6 where it really says $L = {L\over 10}$ (top right). I understand what you mean but it really isn't right. And it's costing marks needlessly.
All this isn't physics, and it barely is math; just common sense.

You can see this 2/3 in your drawing.

If the string has a 3rd harmonic frequency and its an open pipe I though using the equation λs=2/3L. For L I used the diameter of the can because its the length.

At least write something like "λ_s=2/3L_s. L_s = the diameter of the can". And yes, that is the length of the string.

On to your post #18. "confusion, not my strength" Nothing wrong with that. That's what being a student of something is about. Was that a reaction to post #17 or did it cross post #17 and come as an afterthought after post #16 ? Because that was bulls eye, spot on, just right, and what have you.

Then: I hope #13 was an answer to my first ... in post #12. 4L is right. A pipe with one end open and one end closed has its lowest resonance at a frequency corresponding with a wavelength of four times its length.

At a closed end the air can't move (zero amplitude of the back-and-forth motions) so the amplitude of the pressure variations is maximum. At an open end the pressure is equal to the pressure in the outside world, i.e. (almost) zero variations in pressure and maximum amplitude in the collective motion of the air molecules (sigh... I find it difficult to formulate it all correctly and still keep it legible ).​

There is a second set of ... in my post # 12 and I want to pick up on that. The problem statement is clear:

The string is tensioned such that the 3rd harmonic frequency of the vibrating string matches the fundamental frequency of the can

Is it clear to you that the tone is equal, i.e. the frequency ? Your post #13 is a little bit puzzling: you don't say explicitly that this f₁ of the can is equal to the f₃ of the string, which worried me.
$\$
[edit]Don't understand. I got to see posts #21 and higher only after posting this. Thought I should reply to post #20, which looks silly when you two were at #31 already and getting along just fine. Sorry, Simon (not the first time either, today!). And if I didn't make a mess of it, perhaps Emmy doesn't have to suffer from my barging in.

At least I didn't spoil anything. Did I miss the ultimate answer, or did I see it come by at some point