D)We can use the same formula as in part A. L = 343 m.s / 466.2 HzL = 0.737m

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In summary: Ok, so you've got the right idea, but there are a few errors in your working. You wrote,466.2 = n (343 / 2L )which implies that2L = n (343 / 466.2)However, this is not true. If you look at the dimensions of the quantities, you'll find that the right-hand side has units of seconds, while the left-hand side has units of meters. Therefore, the entire equation is dimensionally incorrect. To correct this, you need to change the 466.2 to a frequency, namely 466.2 Hz. This is easily done, since we know that f_n = 466.2 and f_n
  • #1
SEEDS
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Homework Statement


One organ pipe open on both ends and one organ pipe open at one end and closed
at the other are tuned on the same fundamental tone with frequency f = 264 Hz.

a)How long is each pipe?
b) Determine the first three harmonics of each pipe.
c) There is an organ pipe open at both ends, where the frequencies of three neighbo-
ring harmonics are 466:2 Hz, 582:7 Hz and 699:2 Hz. To which harmonics do this
frequencies belong?
d) How long is this pipe?
Velocity c = 343 m/s.whats is the basic difference if an organ pipe in open on both ends or one end closed ?
To find the length in part d) should i apply the same relationship ?

Homework Equations


The Attempt at a Solution



I tried like ,A)
Since Wavelength * Freq = Speed
Therefore Wavelength = Speed / Freq
Wavelength = 343 m.s / 264 Hz
Wavelength =1.299m

For Open Pipe:
L = 1/2 (wavelength)
L = 0.649 meter

For Closed Pipe:
L= 1/4 Wavelength
L = 0.3247 meter
B) Open Pipe Closed Pipe
L = 1/2 Wavelength L = 1/4 Wavelength
L= 2/2 Wavelength L = 3/4 Wavelength
L= 3/2 Wavelength L = 5/4 Wavelength C) Should I apply fn = n f1 if yes so is the f1 = 264 Hz ?
 
Last edited:
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  • #2
Closed and open pipes have different fundamental frequencies. The equation that you used are for open pipes only (L=λ/2), since a closed pipe has a different fundamental frequency you are required to use a different equation to determine the length of the pipes.
 
  • #3
Please don't edit your post once someone has replied. Doing so can make their reply seem wrong or confusing, as in this case.

Which equations are you using for (c)?
 
  • #4
Ohh Sorry !

For C Please take a look at my first post.
 
  • #5
Hootenanny said:
Please don't edit your post once someone has replied. Doing so can make their reply seem wrong or confusing, as in this case.
SEEDS said:
Ohh Sorry !

For C Please take a look at my first post.
Yet you've done it again :grumpy:

f = n*f1 looks like the right equation to apply.
 
  • #6
could you confirm me about the fundamental fr in (C)
 
  • #7
SEEDS said:
could you confirm me about the fundamental fr in (C)
f1 is given to you in the question:
SEEDS said:
One organ pipe open on both ends and one organ pipe open at one end and closed
at the other are tuned on the same fundamental tone with frequency f = 264 Hz.
 
  • #8
If I am using
fn = n f1 so i am getting fractional figures

466.2 / 264 = n and i guess n should'nt be fractional .
582.7 / 264 = n
699.2 / 264 = n
 
  • #9
SEEDS said:
If I am using
fn = n f1 so i am getting fractional figures

466.2 / 264 = n and i guess n should'nt be fractional .
582.7 / 264 = n
699.2 / 264 = n
Hmm, indeed you are correct. Therefore, we must conclude that we are dealing with a different pipe to the one in the previous question. So, let us start again.

What is the general equation for the fundamental frequency of an open pipe?
 
  • #10
Hootenanny said:
What is the general equation for the fundamental frequency of an open pipe?



The fundamental frequ. of an open pipe is

f0 = v/(2*L1)
 
  • #11
SEEDS said:
The fundamental frequ. of an open pipe is

f0 = v/(2*L1)
Correct, so using this equation and your previous one we can write

[tex]f_n = n\frac{v}{2L}[/tex]

Can you go from here?

HINT: You know that the harmonics are consecutive.
 
  • #12
[tex]f_n = n\frac{v}{2L}[/tex]
I am applying it like

n = (fn * 2 L) / v but
n1=2.718 L
n2=3.397 L
n3=4.076 Lstill confusion how to find L to proceed finally for the harmonics ?
 
Last edited:
  • #13
SEEDS said:
[tex]f_n = n\frac{v}{2L}[/tex]



I am applying it like

n = (fn * 2 L) / v but
n1=2.718 L
n2=3.397 L
n3=4.076 L


still confusion how to find L to proceed finally for the harmonics ?
Try writing them in the form

[tex]f_n = n\frac{v}{2L}[/tex]

[tex]f_{n+1} = \left(n+1\right)\frac{v}{2L}[/tex]

[tex]f_{n+2} = \left(n+2\right)\frac{v}{2L}[/tex]
 
  • #14
ok let's check it

fn = n v/2L ---A

fn+1 = (n+1) v/2L ----B

fn = 466.2 Hz , fn+1 = 582.7 Hz

Therefore A becomes

466.2 = n (343 / 2L )
2.718L = n ------(i)

and B becomes by fn+1
582.7L = 171.5 n + 171.5 ---(ii)
putting n from i

L = 1.471 meter
and n is approx = 4
So the harmonics are 4th , 5th , 6th

Is this correct ?
 

1. How did you determine the value of 343 m/s for the speed of sound?

The value of 343 m/s for the speed of sound is considered a standard value that has been experimentally determined through various scientific studies and measurements. It is based on the conditions of temperature, pressure, and humidity in the Earth's atmosphere.

2. What does the unit "m/s" mean in this formula?

The unit "m/s" stands for meters per second and is used to represent the speed of sound, which is the distance traveled by sound waves in one second.

3. Can this formula be used for any frequency of sound?

Yes, this formula can be used for any frequency of sound as long as the speed of sound remains constant. However, it is important to note that the value of 343 m/s is an approximation and may vary slightly depending on the conditions of the environment.

4. How does temperature affect the speed of sound?

The speed of sound is directly proportional to the temperature of the medium it is traveling through. This means that as temperature increases, the speed of sound also increases. This is because molecules in a warmer medium have more energy and can vibrate faster, allowing sound waves to travel faster.

5. Why is the speed of sound different in different mediums?

The speed of sound is affected by the properties of the medium it is traveling through, such as temperature, density, and elasticity. For example, sound travels faster in solids than in liquids and even slower in gases. This is because the molecules in solids are closer together and can transmit sound waves more efficiently.

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