Finding the harmonics of a flute

[Greywolfe1982]

Homework Statement

A flute is essentially a pipe open at both ends.
The length of a flute is approximately 66.3 cm.
The speed of sound in the flute is 342 m/s.
What is the first harmonic of a flute when
all keys are closed, making the vibrating air
column approximately equal to the length of
the flute? Answer in units of Hz.

What is the second harmonic? Answer in
units of Hz.

What is the third harmonic? Answer in units
of Hz.

Homework Equations

I found the equation for fundamental frequency in an open-ended tube online: f=v/2L.

The Attempt at a Solution

f=v/2L
f=342/2x.633
f=270.142

However, my assignment says this is wrong. I searched some more and found that harmonics are multiples of the fundamental frequency (f, 2f, 3f, etc.). I'm sure this is relevant, I'm just not sure how - finding 2f gave me an incorrect answer as well.

 

[Dr.D]

Look at the units of your equation:
On the right side you have v/(2*L) [=] (L/T)/(L) [=] T
so the answer comes out in "per second". What does this mean? It means that the frequency you have calculated is in radians/sec, not in Hz. You need to then adjust the value to put it into Hz.

Does anybody teach unit checking these days? I sure see a lot of unit errors!

 

[Greywolfe1982]

How do you convert something from radians/second to oscillations/second? I understand how the units cancel, but I'm still unsure of what to do to get the correct answer.

 

[BobMonahon]

2 PI radians = 1 oscillation.

So if you have an answer in Radians / second (R/sec), convert to Oscillations / second (Osc/sec) by multiplying:

(R/sec) . (Osc / Radian) = Osc / sec.

In this case, you'll get your answer times (1 Osc / 2 PI Radians) -- or: Your answer divided by 2 PI. Do you get the right answer now?

 

[Greywolfe1982]

Nope, not the right answer =\

270.142/6.283185307=42.9944. Tried that and it wasn't correct. Any ideas on what I'm doing wrong? Did I mess up multiplication somewhere?

 

[LowlyPion]

The Attempt at a Solution

f=v/2L
f=342/2x.633
f=270.142

However, my assignment says this is wrong. I searched some more and found that harmonics are multiples of the fundamental frequency (f, 2f, 3f, etc.). I'm sure this is relevant, I'm just not sure how - finding 2f gave me an incorrect answer as well.

Try recalculating with the .663 that is given in the problem.

 

[Greywolfe1982]

...I hate myself -.- I tend to go through my problems and look over them, rather than rewrite them. Unfortunately this can make me miss stupid mistakes like that.

Thanks a lot for the help.

 

[lokistar]

flutes are closed ended columns so the tube is equal to 1/4 of the wavelength so it would be

4X66.3 in metres so 4X0.663 = 2.652 that is the wavelength

to find the frequency it is speed/wavelength which is 129.7hz so assuming variants in atmospheric pressure will occur this frequency can be reguarded as C which is 130hz with a wavelength of 264cm
so to check 264/4=66.3 the length of the tube.

so the harmonics are 1st 130hz
2nd 260hz
3rd 390hz
4th 520hz

i think that's right I am pretty sure that's how it goes