Problem Related to Beat Frequency of a hollow tube

Click For Summary
SUMMARY

The discussion focuses on calculating the beat frequency produced by a hollow metallic tube closed at one end when subjected to a temperature increase. The relevant equations include the frequency formula for a closed tube, n = v/(4l), and the relationship between frequency, wavelength, and the speed of sound, n = c/λ. Participants emphasize the need to account for the coefficient of thermal expansion (alpha) and the changes in both the speed of sound and the length of the tube due to temperature variations. The challenge lies in expressing the beat frequency in terms of temperature (T) and alpha, while maintaining dimensional consistency.

PREREQUISITES
  • Understanding of wave mechanics, specifically sound waves.
  • Familiarity with the concept of resonance in closed tubes.
  • Knowledge of thermal expansion and its effects on material properties.
  • Ability to manipulate equations involving frequency, wavelength, and speed of sound.
NEXT STEPS
  • Research the relationship between temperature and the speed of sound in air.
  • Study the effects of thermal expansion on the dimensions of materials.
  • Learn how to derive beat frequency equations in acoustics.
  • Explore the concept of resonance in different types of tubes and their applications.
USEFUL FOR

Students studying physics, particularly those focusing on acoustics and thermodynamics, as well as educators seeking to explain the principles of sound resonance and thermal effects on materials.

digitomega
Messages
3
Reaction score
0

Homework Statement



Q. A hollow metallic tube closed at one end produces resonance with a tuning fork of frequency n at temperature T. The temperature of the tube is increased by dT. If the coefficient of thermal expansion of the tube metal is alpha, the beat frequency will be - (there are 4 option in the attached image)

Homework Equations


Frequency for a tube closed at one end - n=v/(4l),3v/(4l)...
[/B]

The Attempt at a Solution


I have no idea ho to get the solution in terms of temperature and coefficient of thermal expansion.
[/B]
 

Attachments

  • IMG_20170303_101912.jpg
    IMG_20170303_101912.jpg
    9 KB · Views: 584
Physics news on Phys.org
I'm not sure I understand the question.
It is a simple matter to find the change in length of the tube and consequent change in frequency, but a beat requires two frequencies at once. Seems to me there will be one frequency at one temperature and a different frequency at the raised temperature.

Doh! Forgot about the tuning fork. Thanks TSny.
 
Last edited:
  • Like
Likes   Reactions: digitomega
I think that when the temperature of the tube is raised, the tuning fork still vibrates at its original frequency. I guess you assume that the fork can still excite the same mode in the tube, but now the mode has a slightly different frequency than the tuning fork.
 
  • Like
Likes   Reactions: digitomega
You are right haruspex. I also tried finding frequency in the two cases but the required answer is in terms of T and alpha. I can get alpha in the answer expression but I have no idea about how to get T in the expression.
 
digitomega said:
I have no idea about how to get T in the expression.
Assume the gas inside the tube is also heated.
 
digitomega said:
You are right haruspex. I also tried finding frequency in the two cases but the required answer is in terms of T and alpha. I can get alpha in the answer expression but I have no idea about how to get T in the expression.
Please post some working...
How does the temperature relate to the length?
How does the wavelength relate to the length?
How does the temperature relate to the speed of sound in the air?
How does the frequency relate to the wavelength and the speed of sound?
 
haruspex said:
Please post some working...
How does the temperature relate to the length?
How does the wavelength relate to the length?
How does the temperature relate to the speed of sound in the air?
How does the frequency relate to the wavelength and the speed of sound?

For a tube closed at one end-
1/λ = v/4l (λ= wavelength and v= speed of sound).
Increasing the temperature ultimately increases speed as well as frequency of sound.
for sound there is the relation n=c/λ. (n=frequency, c=speed of sound,λ=wavelength)
As for length and temperature relation, increasing temperature increases length.
24b.gif


Problem is I don't know and can't find how to replace l and v in the expression I am getting for the beat frequency to make it like one of the 4 options.
 
digitomega said:
1/λ = v/4l (λ= wavelength and v= speed of sound).
Something wrong there. It doesn't make sense dimensionally.
digitomega said:
Increasing the temperature ultimately increases speed as well as frequency of sound.
Well, it changes the speed of sound. It does not directly change the frequency. What is the equation?

Once you have the right relationship in each case, you need either to write a second copy of each, using different symbols for the changing variables (e.g. T for initial temperature, T' for final temperature, etc.) or write the "delta" equations, e.g. how the change in temperature, ΔT, relates to the change in speed of sound, Δc.
 

Similar threads

Replies
4
Views
3K
  • · Replies 14 ·
Replies
14
Views
2K
Replies
3
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
1
Views
3K
  • · Replies 7 ·
Replies
7
Views
6K
Replies
1
Views
3K
  • · Replies 43 ·
2
Replies
43
Views
7K
  • · Replies 6 ·
Replies
6
Views
5K
  • · Replies 7 ·
Replies
7
Views
2K