Resonance Column Method

[chaoseverlasting]

Homework Statement

In the experiment of the determination of the speed of sound in air using the resonance column method the length of the air column that resonantes in the fundamental mode with a tuning fork is 0.1m. When this length is changed to 0.35m, the same tuning fork resonates with the first overtone. Calculate the end correction.


In a resonance tube, a tuning fork of 512 hz resonates with the tube. The first resonance gives a 30.7cm rise of water level and the second one gives 63.2 cm rise. Calculate the error in calculating the velocity of sound in air.

Homework Equations

$$\nu = (2n-1)v/4L or \nu =nv/2L$$

The Attempt at a Solution

I don't know what the end correction is or what the resonance air column method is. Is the pipe open at both ends or at just one end? And what do they mean by end correction. If someone could explain the experiment to me, I could probably solve the question.

 

[Mentz114]

I think you should describe the experiment. You don't seem to have made much effort.

 

[AlephZero]

In the first question, the second resonant length is about 3 times the first length. In the second question, it is about 2 times the first length. That tells you whether the tubes were open or closed in each case.

The "end correction" means that the length of the resonating air column is not exactly the same as the length of the tube. It equals L+C where L is the length of the tube and C is the correction. The resonant frequencies are always exactly the same as the tuning fork frequency.

I don't understand exactly what they want you to do in the second question.

 

[nrqed]

Homework Statement

In the experiment of the determination of the speed of sound in air using the resonance column method the length of the air column that resonantes in the fundamental mode with a tuning fork is 0.1m. When this length is changed to 0.35m, the same tuning fork resonates with the first overtone. Calculate the end correction.


In a resonance tube, a tuning fork of 512 hz resonates with the tube. The first resonance gives a 30.7cm rise of water level and the second one gives 63.2 cm rise. Calculate the error in calculating the velocity of sound in air.

Homework Equations

$$\nu = (2n-1)v/4L or \nu =nv/2L$$

The Attempt at a Solution

I don't know what the end correction is or what the resonance air column method is. Is the pipe open at both ends or at just one end? And what do they mean by end correction. If someone could explain the experiment to me, I could probably solve the question.

It's a tube partialy filled with water. One end is open to air with a source of sound placed near the opening. The setup is such that one may vary the level of the water thereby changing the length of the air column (which is basically the distance between the water level and the opening of the tube).

The idea is then to keep the frequency fixed to some value and to adjust the level of the water until one hears a resonance. In *theory*, the first resonance (corresponding to the the water level being the closest to the opening of the tube hence the smallest length of the air column) should be at a quarter of a wavelength. The distance between each resonance should be half a wavelength. In practice, the distance between adjacent resonance is indeed half a wavelength but the first resonance is not exactly at a quarter of wavelength (because the antinode is not exactly at the rim but extends slightly outside of the opening of the tube). The difference between the actual length of the air column for the first resonance and lambda/4 is what they call the "correction".

Hope this helps

Patrick