Coke bottle on another planet, involving harmonics and speed

In summary, the problem involves finding the speed of sound on Earth 2 given a frequency of 520 Hz and a length of 0.179 m for a closed-end cylinder. The second available harmonic for this frequency is used in the equation v = (4/3)Lf to find the speed, resulting in an answer of approximately 41.3 m/s.
  • #1
riseofphoenix
295
2
Number4-1.png


This is what I tried doing, but I ended up getting an answer that isn't listed above

1. frequency = speed/wavelength

2. On Earth

480 Hz = (343 m/s)/wavelength
wavelength = 343 / 480
wavelength = 0.714 is the wavelength for the Coke bottle

3. On Earth 2

Since they already gave me frequency on Earth 2, and I already solved for wavelength, all I have to do is plug in those numbers to solve for speed on Earth 2.

frequency = speed/wavelength
(520 Hz) = speed/0.714
(0.714)(520) = speed
371.58 m/s = speed
 
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  • #2
Wait...I think first i have to find the harmonics...
 
  • #3
You're dealing with a fundamental frequency on Earth, but a harmonic on Earth 2. What harmonics can a closed-bottom cylinder produce?
 
  • #4
gneill said:
You're dealing with a fundamental frequency on Earth, but a harmonic on Earth 2. What harmonics can a closed-bottom cylinder produce?

n = 1, 3, 5, 7

So...

On Earth 2:

520 = 5*104 (fifth harmonic of 104 Hz)
 
  • #5
gneill said:
You're dealing with a fundamental frequency on Earth, but a harmonic on Earth 2. What harmonics can a closed-bottom cylinder produce?

You still there?
 
  • #6
riseofphoenix said:
You still there?

Still here. You should confirm which harmonics can be produced by a closed-end pipe.
 
  • #8
So, when the problem states that "a second available harmonic of 520Hz is available", which one do you think they're referring to?
 
  • #9
gneill said:
So, when the problem states that "a second available harmonic of 520Hz is available", which one do you think they're referring to?

312/3 = 3*104 (second available harmonic of 104)
520/5 = 5*104 (third available harmonic of 104)
 
  • #10
riseofphoenix said:
312/3 = 3*104 (second available harmonic of 104)
520/5 = 5*104 (third available harmonic of 104)

Why 104? The frequencies will go: 520, 520/3, 520/5, 520/7,...

The second one is 520/3, right?
 
  • #11
gneill said:
Why 104? The frequencies will go: 520, 520/3, 520/5, 520/7,...

The second one is 520/3, right?

Ohh!

that's right, because I was thinking the whole time the the numbers HAD to be integers...

Yeah that's what I meant to say!

But what next though?Do I then do:

1. frequency = speed/wavelength

2. On Earth

480 Hz = (343 m/s)/wavelength
wavelength = 343 / 480
wavelength = 0.714 is the wavelength for the Coke bottle

3. On Earth 2

Since they already gave me frequency on Earth 2, and I already solved for wavelength, all I have to do is plug in those numbers to solve for speed on Earth 2.

frequency = speed/wavelength
(173.3 Hz) = speed/0.714
(0.714)(173.3) = speed
123.76 m/s = speed?
 
Last edited:
  • #12
gneill said:
Why 104? The frequencies will go: 520, 520/3, 520/5, 520/7,...

The second one is 520/3, right?

Is 123.9 the answer to this problem?
 
  • #13
riseofphoenix said:
Is 123.9 the answer to this problem?

You'll have to show your work; I can't verify a guess :smile:
 
  • #14
gneill said:
You'll have to show your work; I can't verify a guess :smile:

I just did :eek:

frequency = speed/wavelength
(173.3 Hz) = speed/0.714
(0.714)(173.3) = speed
123.76 m/s = speed
 
  • #15
I have a problem with the length of column (bottle) that you're using (Think about a coke bottle over 70cm tall!). While the error has fortunately canceled out along the way, I can't in good conscience say that your answer is altogether correct :smile:

On the web page that you found, take a look at the table partway through that shows the column length \ wavelength relationship for the various harmonics. Note that the fundamental is NOT the same length as the column.
 
  • #16
gneill said:
I have a problem with the length of column (bottle) that you're using (Think about a coke bottle over 70cm tall!). While the error has fortunately canceled out along the way, I can't in good conscience say that your answer is altogether correct :smile:

On the web page that you found, take a look at the table partway through that shows the column length \ wavelength relationship for the various harmonics. Note that the fundamental is NOT the same length as the column.

What what??

You mean

Wavelength = (4/3)*L
Wavelength = (4/3)*(0.714)
Wavelength = 0.952?
 
  • #17
I'm so confused now...
 
  • #18
On Earth 1, the fundamental is 480Hz, speed 343m/s. Wavelength is then 0.715m. That's fine. But the bottle has length L = λ/4, so L = 0.179m. The length is assumed not to vary between planets (same bottle).

So take that bottle length and the harmonic 520Hz and use the appropriate length versus wavelength relationship to find the speed (since v/f = λ, and λ = (4/3)L for this harmonic).

Sorry, got to go now to an important meeting with a pint of something refreshing :wink:
 
  • #19
Earth 1

Step 1) Find wavelength
Fundamental frequency on Earth 1: 480 Hz
Speed on Earth 1: 343 m/s.
Wavelength on Earth 1: λ = v/f, which is 343/480 = 0.715 m

Step 2) Find the length of the bottle
The bottle has length L = λ/4
L = (0.715 m)/4
L = 0.179 m

The length is assumed not to vary between planets (same bottle).

Earth 2

Step 3)
Take that bottle length (L = 0.179) and find the second available harmonic of 520 Hz.

Since the bottle is a close-ended cylinder, n = 1, 3, 5, 7 and the frequencies will go: 520, 520/3, 520/5, 520/7. The second available harmonic of 520 Hz is 520/3 = 173.3
According to: http://www.physicsclassroom.com/class/sound/u11l5d.cfm

Step 4) To find speed, use v = λf (speed = wavelength*frequency). λ for the second available harmonic of 520 Hz will = (4/3)L

λ = (4/3)(0.179)
λ = 0.239

So,

v = λf
v = (0.239)(173.3)
v = 41.3 m/s
 
  • #20
Thanks!
 
  • #21
Hmm. If L = 0.179m and f = 520 Hz for the harmonic n = 3 where n = 1, 3, 5, 7,... then

## v = \frac{4}{3}L f = 123.9 m/s ##
 
  • #22
gneill said:
Hmm. If L = 0.179m and f = 520 Hz for the harmonic n = 3 where n = 1, 3, 5, 7,... then

## v = \frac{4}{3}L f = 123.9 m/s ##

Wait whattttt??

I already submitted it :frown:

Are you sure that's the answer?

I plugged in what you just wrote and got another answer.

v = (4/3)(0.179)(173.3)
v = 0.238666667(173.3)
v = 41.3?

If that's not correct and if you just plugged in f = 520 Hz, it gives a completely different answer (?)

v = (4/3)(0.179)(520)
v = 0.238666667(520)
v = 21507.68533?

What did you plug in??
 
  • #23
On Earth 2:

For the second available harmonic, harmonics being 1, 3, 5, 7,...

##v = f \frac{4}{3}L##

Where f = 520 Hz, L = 0.179 m.
 
  • #24
gneill said:
On Earth 2:

For the second available harmonic, harmonics being 1, 3, 5, 7,...

##v = f \frac{4}{3}L##

Where f = 520 Hz, L = 0.179 m.

-.-
kjbllajds... fudge.

ok.
thanks
 

Related to Coke bottle on another planet, involving harmonics and speed

1. How would the harmonics of a Coke bottle on another planet differ from those on Earth?

The harmonics of a Coke bottle on another planet would depend on the atmospheric conditions and the composition of the planet's surface. If the planet has a different atmosphere or composition than Earth, the sound waves produced by the bottle would be affected, resulting in different harmonics.

2. Could the speed of sound on another planet impact the resonance of a Coke bottle?

Yes, the speed of sound on another planet could impact the resonance of a Coke bottle. The speed of sound is affected by factors such as temperature, pressure, and density of the medium it is traveling through. If these factors are different on another planet, the speed of sound would also be different, affecting the resonance of the bottle.

3. Would a Coke bottle produce the same sound on a planet with no atmosphere?

No, a Coke bottle would not produce the same sound on a planet with no atmosphere. Sound waves require a medium to travel through, and without an atmosphere, there would be no medium for the sound waves to travel through. This would result in no sound being produced by the bottle.

4. How would the size and shape of a Coke bottle impact its harmonics on another planet?

The size and shape of a Coke bottle can impact its harmonics on another planet. The larger the size of the bottle, the lower the frequency of the sound waves produced. Similarly, the shape of the bottle can affect the resonance and harmonics produced. Different shapes will produce different harmonics and frequencies.

5. Can the speed of a Coke bottle's vibrations be altered by the gravity on another planet?

Yes, the speed of a Coke bottle's vibrations can be altered by the gravity on another planet. Gravity affects the speed at which objects vibrate, so on a planet with a different gravity than Earth, the vibrations of the Coke bottle would also be different.

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