# Help regarding standing waves

[SOLVED] Help regarding standing waves

a.The flute player is very musical and produces a pure tone of frequency f=600Hz. For the flute to be in 1st harmonic, what is d where d is the length of the flute?

b.By How much has d to be increased to increase the number of nodes by 2? (And keep the frequency the same)

c.What frequency would be necessary to obtain the first harmonic at the piston in the position calculated in previous part?

a) d=.14291meters
b)d increased by .57167 meters
c) frequency is equal to 120 Hz

Would anyone be able to check those answers? Thank you!

I have figured out that the first solution is correct. May anyone verify the rest?

Anyone?

dynamicsolo
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a.The flute player is very musical and produces a pure tone of frequency f=600Hz. For the flute to be in 1st harmonic, what is d where d is the length of the flute?

b.By How much has d to be increased to increase the number of nodes by 2? (And keep the frequency the same)

c.What frequency would be necessary to obtain the first harmonic at the piston in the position calculated in previous part?

a) d=.14291meters

I gather that the value being used for the speed of sound is
343 m/sec. The flute is being treated as a half-closed pipe, so its fundamental frequency (first harmonic) corresponds to a standing wavelength which is 4 times the length of the flute (or L = lambda/4).

b)d increased by .57167 meters

Increasing the number of nodes by 2 means that two half-wavelengths more must be contained within the pipe. So the pipe must be lengthened by one full wavelength, which is
(343 m/sec)/(600 Hz).

c) frequency is equal to 120 Hz

You can get this in one of a couple of ways. One is that the pipe is now 0.7146 m. long, which is one-fourth of the new wavelength (for the new fundamental frequency); thus, that wavelength is 2.8584 m., corresponding to a frequency of (343 m/sec)/2.8584 m. The other approach is to note that the new pipe is five times the length of the original flute, so the new wavelength for the first harmonic must be five times longer than the original first harmonic, making the new fundamental frequency (1/5) · 600 Hz.

Last edited:
alphysicist
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Hi yahoo32,

How did you verify your answer for part a? A flute is a pipe open at both ends, but I think your answer for part a is for a pipe with one end closed and one end open.

Thanks a lot! The final question asks:

THe flute player now produces a tone with frequency f = 300 H z (in the ﬁrst harmonic).
How long is the ﬂute ?

This is just the same as part a) but with a different frequency right?

dynamicsolo
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Ah, I was assuming that these answers were being compared to those from a text. But, yes, upon checking around, it seems that a flute should be treated as a pipe with both ends open. What is the relation of the fundamental wavelength to the length of the pipe in that case?

No, the answers were those that I had solved for. However, I think it is one end open and one end closed because the beginning of the problem states

The musician is well known for his flute; the piston can be
moved such that the length of the air column d can be adjusted. Assume the
velocity of sound in air is 343 m/s.

Thanks again!

dynamicsolo
Homework Helper
Thanks a lot! The final question asks:

THe flute player now produces a tone with frequency f = 300 H z (in the first harmonic).
How long is the flute ?

This is just the same as part a) but with a different frequency right?

That's true, but now all of your answers need to be revised, applying the result for the fundamental wavelength of a fully-open pipe. (The flute length changes in part (a) and the multiplier for the other parts is no longer 5...)

alphysicist
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Hi yahoo32,

No, the answers were those that I had solved for. However, I think it is one end open and one end closed because the beginning of the problem states

The musician is well known for his flute; the piston can be
moved such that the length of the air column d can be adjusted. Assume the
velocity of sound in air is 343 m/s.

Thanks again!

That's why it is important to post the entire question! To many people (at least in the US, I suppose; I don't know about elsewhere) a flute is an instrument with both ends open. The instrument that the problem refers to would I think (again, here in the US) commonly be called a slide whistle.

But the problem wording itself indicates that the pipe has one end closed by a piston, so the whole issue of names would not have mattered when we can read the entire question.

dynamicsolo
Homework Helper
The musician is well known for his flute; the piston can be
moved such that the length of the air column d can be adjusted. Assume the
velocity of sound in air is 343 m/s.

OK, so what alphysicist is pointing out about flutes is correct, but the instrument described in the problem does not function like a flute, but more like, say, an organ pipe. So if that's what the problem intended, I guess you're all set...

Ahh! I see so now the flute is open therefore the fundamental frequency is equal to v/2L . So if the frequency of the first harmonic is 300Hz, L should .5716m correct? (I am referring to the last part of the problem)

dynamicsolo
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So if the frequency of the first harmonic is 300Hz, L should .5716m correct? (I am referring to the last part of the problem)

For the fully-open flute, yes; and the 600 Hz flute would be then half that length (referring back to part (a) ).

alphysicist
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We seemed to have posted posts #10, 11, and 12 all within a minute of each other.

Yahoo32: in post #10 I explained how I became confused. I think the answers in your original post are correct, and you do part d the same way you did part a in the original post.

THank you so much for your help guys! I had one last questions regarding standing waves in general:

If you have a a standing wave on a string with 2 closed ends and it has 2 anti nodes and 3 nodes (if you count the ends). If the Tension is divided by 4 while the frequency and the length are constant, the new standing wave produced is the first harmonic, correct ? (With 1 anti node and 2 nodes if you count the ends)

alphysicist
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THank you so much for your help guys! I had one last questions regarding standing waves in general:

If you have a a standing wave on a string with 2 closed ends and it has 2 anti nodes and 3 nodes (if you count the ends). If the Tension is divided by 4 while the frequency and the length are constant, the new standing wave produced is the first harmonic, correct ? (With 1 anti node and 2 nodes if you count the ends)

I don't think that sounds right; can you show how you got that answer? In particular, what happens to the speed of the wave?

Essentially I used the equation f(sub n) = n/2L *square root of (T/ (mu)). My n is equal to 2 since there are 2 anti nodes and I chose arbitrary numbers for T, mu, and L and got a frequency. I did the same process again but i divided my original T by 4 and found that the new frequency is exactly half that of the first. Therefore I concluded that the new wave must be at the first harmonic? With one anti node and the 2 nodes (one at each end).

alphysicist
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But I think the problem indicates that the frequency is held constant; it is the wavelength and speed that will change.

Hmmm... you are correct. With my method my frequency i snot held constant. How would I figure out how many nodes and anti nodes are on this wave now?

alphysicist
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I would first draw the original wave; if the length of the string is L, what is the wavelength?

Then find out how the velocity will change if you reduce the tension by a factor of 4. Once you have that, what is the new wavelength (using v=f(lambda) )? If you draw a picture of the new string, when the same length L and the new wavelength, you can count the nodes and therefore tell which harmonic it is. What do you get?

for the original wave the wave length would be L. And the velocity will be halved if I reduce the tension by a factor of 4. I didn't get your last part. Would the new wave length just be v/2f?

alphysicist
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That's right, but I think you want it in term of L?

original wavelength: $$\lambda_i = L = \frac{v}{f}$$

new wavelength: $$\lambda_f = \frac{v}{2f}$$

Once you know how many wavelengths will fit along your string, what do you get for the harmonic?

Ahh so now it is 2L. Therefore you would have 4 anti nodes and 5 nodes. Is that correct?

alphysicist
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Ahh so now it is 2L. Therefore you would have 4 anti nodes and 5 nodes. Is that correct?

Well, it's L/2. But, yes, there are 4 anti-nodes, because you can fit two complete waves along the string.

Wait, really? Now I am confused. If you fit in L/2 Wavelengths how do you get 4 anti nodes?

alphysicist
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The new wavelength is L/2. So if the string is 1 meter long, each wave is only 0.5 meters and there is room for two waves, each with two antinodes.

I see. Thanks a lot for your help, it really makes sense now!

alphysicist
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