Resonant frequency and wire tension in closed tube with oscillating wire

[ghetto_bird25]

Homework Statement

A tube 1.03 m long is closed at one end. A stretched wire is placed near the open end. The wire is 0.37 m long and has a mass of 8.2 g. It is fixed at both ends and oscillates in its fundamental mode. By resonance, it sets the air column in the tube into oscillation at that column's fundamental frequency. Find (a) that frequency and (b) the tension in the wire. (Take the speed of sound in air to be 343 m/s.)

Homework Equations

well in this particular question i think you have to use the formula f=$$\frac{v}{\lambda}$$
in which case lamda is equal to 4L/2n+1 and then you find the derivative of that set it equal to zero and find your n
plug that back in and i think you mite get frequency but i keep getting a negative n value

The Attempt at a Solution

i guess u read above

 

[andrevdh]

The wavelength of the fundamental resonance tone of the tube forms with n = 0. That is a quarter of a wavelength of the oscillation will fit into the tube. The stationary point is at the closed end and the maximum displacement of the air molecules is at the open end.

 

[ghetto_bird25]

so the frequency would be zero?

 

[andrevdh]

No, you said that

"... in which case lamda is equal to 4L/2n+1 ..."

that is the requirement for uneven amount of quarter wavelengths to fit into the tube. Why uneven quarters? A node forms at the closed end and an antinode at the open end.