Open-Closed Tube: Tension in 24.0cm Wire at Fundamental Frequency

[skinard360]

Here is my question:

A 24.0 cm -long wire with a linear density of 20.0 g/m passes across the open end of an 89.0 cm-long open-closed tube of air. If the wire, which is fixed at both ends, vibrates at its fundamental frequency, the sound wave it generates excites the second vibrational mode of the tube of air. What is the tension in the wire? Assume v=340 m/s

I am confused about the two values for the length. Do I have to use two equations and set them equal to each other? Would all else be the same for both? Please help

 

[denverdoc]

yes you do need to use both lengths, what the systems have in common is frequency where the pipe is in its second vibrational mode, while the string is at its fundamental. Do you know what second mode means here?

so write down some eqns, for both systems and we can go from there.

 

[m2003]

I can provide some clarification on the given information. The length of the wire and the length of the open-closed tube are two separate values. The wire has a length of 24.0 cm and the open-closed tube has a length of 89.0 cm. These two lengths are not directly related and do not need to be set equal to each other.

To calculate the tension in the wire, we can use the formula T = (m/L)v^2, where T is the tension, m is the linear density, L is the length, and v is the speed of the sound wave in air (given as 340 m/s).

Substituting the given values, we get:
T = (20.0 g/m * 0.24 m) * (340 m/s)^2
T = 163.2 N

Therefore, the tension in the wire is approximately 163.2 N. The length of the open-closed tube does not affect this calculation, as it is only related to the frequency of the sound wave and the mode of vibration in the tube.

I hope this helps clarify any confusion and please let me know if you have any further questions.