Solve for Frequency and Velocity in a Uniform Tube with Open Ends

[gaobo9109]

Homework Statement

A uniform tube, 60.0cm long, stands vertically with its lower end dipping into water. When the length above water is 14.8cm, and again when it is 48.0cm. the tube resounds to a vibrating fork of frequency 512Hz. Find the lowest frequency to which the tube will resound when it is open at both ends

Homework Equations

The Attempt at a Solution

When the tube is open at both end, the fundamental frequency f₁ is given by:
f₁=v/(2x0.6)
From the given condition
512 = （2n₁-1)v/(4x0.148)
512 = (2n₂-1)v/(4x0.48)

How can I find the velocity from the abve two equations?

 

[Chi Meson]

These problems can be a lot easier if you draw a diagram of tube resonance (using transverse waves, nearly all textbooks do this). If you keep the medium and frequency the same, and increase the length of the tube, you will see that the difference in tube length between any two successive harmonics is half a wavelength. This is true for open and closed tubes.

 

[Swap]

As Chi Meson stated "difference in tube length between any two successive harmonics is half a wavelength", using this u can easily find the wavelength and then u can find velocity. U will find that the lowest frequency for the tube will be in the second harmonic.

 

[Swap]

Btw is the answer 566.7 Hz ??

 

[Chi Meson]

As Chi Meson stated "difference in tube length between any two successive harmonics is half a wavelength", using this u can easily find the wavelength and then u can find velocity. U will find that the lowest frequency for the tube will be in the second harmonic.

The lowest frequency will be the fundamental, or first harmonic.

The data for the closed-end tube is taking the "end-correction" into account. The length of the tube for the first harmonic should be 1/4 wavelength, and ideally this should be the same as half the difference of the tube lengths as mentioned above. These numbers are not the same because the point of reflection is actually some distance beyond the tube end. Ths distance is determined by the geometry and width/diameter of the tube.

I would apply the end-correction to both ends of the open tube to determine the fundamental wavelength that resonates in it.