What is the velocity of standing waves in a pipe based on resonance distances?

Click For Summary
SUMMARY

The velocity of standing waves in a pipe can be calculated using the resonance distances and the frequency of the tuning fork. In this discussion, a tube filled with air at 770°C and a tuning fork oscillating at 500 Hz produced resonance at distances of 18 cm, 55.5 cm, and 93 cm. The calculated wavelengths were 72 cm, 74 cm, and 74.4 cm, respectively, leading to an approximate wave velocity of 360 m/s. The discussion highlights the importance of considering the "end effect" at the open end of the tube, which can slightly alter the effective length and thus the calculated velocity.

PREREQUISITES
  • Understanding of wave mechanics and resonance
  • Familiarity with the wave equation: λ = u/f
  • Knowledge of sound wave behavior in different mediums
  • Basic principles of acoustics and tube harmonics
NEXT STEPS
  • Study the effects of temperature on sound velocity in gases
  • Learn about the "end effect" in acoustics and its implications
  • Explore the calculation of harmonics in open and closed tubes
  • Investigate the relationship between frequency, wavelength, and velocity in different mediums
USEFUL FOR

Students studying physics, particularly those focusing on wave mechanics, acoustics, and resonance phenomena in tubes. This discussion is also beneficial for educators seeking practical examples of wave calculations in real-world scenarios.

Karol
Messages
1,380
Reaction score
22

Homework Statement


A tube is filled with air at 770C, one end open and on the other a piston. in the open end a tuning fork oscillates with 500[Hz].
The piston is set to different positions and at the distances of 18, 55.5 and 93 cm from the open end there is resonance. find the velocity of the waves.

Homework Equations


\lambda=wave length, u=velocity: \lambda=\frac{u}{f}

The Attempt at a Solution


I understand only one wave length is created, otherwise there will be different velocities.
For the shortest distance: \lambda=4\cdot 18[cm]=72[cm]
For the middle distance: \lambda=\frac{4}{3}\cdot 55.5[cm]=74[cm]
And for the longest: \lambda=\frac{4}{5}\cdot 93[cm]=74.4[cm]
I took the approximate mean of these wavelength: 0.73[m]=\frac{u}{500}\rightarrow u=360
Is that correct?
 
Physics news on Phys.org
There is usually an "end effect" at the open end, which slightly changes the effective length. So consider the tube to be effectively longer by some constant x in each case. You will find that this gives a quite consistent result (and slightly larger than you have calculated).
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
3K
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 33 ·
2
Replies
33
Views
4K
  • · Replies 5 ·
Replies
5
Views
6K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 15 ·
Replies
15
Views
2K