Grade 11 musical instrument project

[ta4294]

Grade 11 musical instrument project need help!

Homework Statement

I am making chime bars for my project.
My partner and I were working on it today and we cut our first chime bar and used visual analyzer to figure out the frequency of sound produced by hitting our first chime bar. We found that it was 410 Hz (using Visual analyzer 2011) (it soudned very close to an A flat when we compared it to an online tuning fork). We the measured the length of our chime bar and it was 42 cm exactly. We then wanted to produce a chime bar that had a frequency of 493.92 ( a B). We cut another piece of pipe of length 35 cm which is what we determined using the F1L1=F2L2 formula . ( (410Hz)(42cm)=(493.92Hz)(L2) , we then solved for L2 which was approximately 34.9 cm) After cutting the new copper pipe at 35cm, we tested its frequency on Visual analyzer 2011. We had used the exact same type of pipe that we had cut our first A flat chime bar from (they both came from the same 6foot sample ) we were holding both of them at their nodes, and we were hitting the chime bar with the same material at their respective antinodes. But the frequency of our 35cm bar came out at 570 Hz which was way off our calculations. I was wondering what might have gone wrong?

Homework Equations

f1l1=f2l2
f=frequency l=length

The Attempt at a Solution

(410Hz)(42cm)=(493.92Hz)(L2)
L2=34.86

Yet chime bar of L2 produces a frequency of 570 Hz not 493.92

 

[I like Serena]

Basically I see 2 possibilities.

1. You used a different pipe (material, thickness, diameter).

2. You made a mistake in the length.
If I calculate the length that corresponds to 570 Hz, I get 30 cm.
Since it was supposed to be 35 cm, that kind of looks like a mistake of exactly 5 cm...

 

[ta4294]

Basically I see 2 possibilities.

1. You used a different pipe (material, thickness, diameter).

2. You made a mistake in the length.
If I calculate the length that corresponds to 570 Hz, I get 30 cm.
Since it was supposed to be 35 cm, that kind of looks like a mistake of exactly 5 cm...

i cut both chime bars (the first one and the second one ) BOTH from the same piece of 6 foot pipe that i bought at the store. I measured my cut piece and it was in fact 34.8 and producing a sound of 570Hz instead of your proposed 30 cm which I agree is what should be the "theoretical value" according to the calcuation. i was looking aroudn the internet and foudn this equation that proposes that L2/L1= square root(f1)/ square root(F2) ... so basically is length inversely proportionate to the root of the frequency? I tried working the numbers out and when L2= 42* ( sqrt(410)/sqrt (570)) <-- (rearranged the aforementioned formula and solved for L2, plugging in the variables that stated in my original problem) i got 35.5! which makes sense because my 35 cm bar was producing 570 Hz... HAVE I BEEN USING THE WRONG EQUATION THE ENTIRE TIME?!

 

[eumyang]

Yep, looks like you've been using the wrong equation. The frequency is inversely proportional to the square of the length of the chime, or
$$F_1 {L_1}^2 = F_2 {L_2}^2$$
which is essentially the 2nd equation you found.

 

[ta4294]

Yep, looks like you've been using the wrong equation. The frequency is inversely proportional to the square of the length of the chime, or
$$F_1 {L_1}^2 = F_2 {L_2}^2$$
which is essentially the 2nd equation you found.

and is this true for like everything? stupid question i know but were learning sound waves and our teacher told us that it was a linear inverse relationship between the frequency and length of something? or is this equation only for chime bars... (i have a feeling its not)

 

[I like Serena]

i cut both chime bars (the first one and the second one ) BOTH from the same piece of 6 foot pipe that i bought at the store. I measured my cut piece and it was in fact 34.8 and producing a sound of 570Hz instead of your proposed 30 cm which I agree is what should be the "theoretical value" according to the calcuation. i was looking aroudn the internet and foudn this equation that proposes that L2/L1= square root(f1)/ square root(F2) ... so basically is length inversely proportionate to the root of the frequency? I tried working the numbers out and when L2= 42* ( sqrt(410)/sqrt (570)) <-- (rearranged the aforementioned formula and solved for L2, plugging in the variables that stated in my original problem) i got 35.5! which makes sense because my 35 cm bar was producing 570 Hz... HAVE I BEEN USING THE WRONG EQUATION THE ENTIRE TIME?!

I've looked it up on wikipedia and found the following article: http://en.wikipedia.org/wiki/Acoustic_resonance.
It seems to me that the ratio between length and frequency is linear.

In the article you can see that an "open" pipe of negligible diameter has a resonance frequency of:

$$f = \frac {n v} {2 L} = \frac {n \times 343 \frac m s} {2 \times 0.42 m} = n \times 408 Hz$$

where n is a positive integer (n=1,2,3,...).

This matches your initial measurement with a chime of length 42 cm and with the "root" tone.

For a length of 35 cm we get:

$$f = \frac {n v} {2 L} = \frac {n \times 343 \frac m s} {2 \times 0.35 m} = n \times 490 Hz$$

which matches what you were trying to achieve.

Is it possible your pipe is for instance "closed" on one end?