A pipe, a string, and resonance

[alexfloo]

Homework Statement

You have a stopped pipe of adjustable length close to a taut 85.0-cm, 7.25-g wire under a tension of 4150*N. You want to adjust the length of the pipe so that, when it produces sound at its fundamental frequency, this sound causes the wire to vibrate in its second overtone with very large amplitude.

How long should the pipe be?

Homework Equations

We are to assume that the speed of sound = 344 m/s.

The Attempt at a Solution

I've taken to typing out all my work for these problems so I'm just going to copy-paste.

l = 0.85 m m = 0.00725 kg
λ = l = 0.85 m for a string in it's second overtone, fixed at both ends.
m = 0.00725 kg

And the wavespeed in the string,
v = sqrt(Tl/m)
= 697.5325972 m/s

And,
v = λf
f = v/λ
= 820.6265849 Hz

Now, the fundamental of the pipe:
λ = 4L for a pipe stopped at one end.

And the wavespeed therein is equal to the speed of sound in air, so
v = λf
f = 344/(4L)
= 86/L

Essentially, we want L such that

86/L = 820.6265849 Hz
L = 86/820.6265849 m
= 0.1047979697 m

There's my answer, but the online homework system says nay. Any ideas?

 

[da_nang]

I believe the error lies in your length-wavelength equation for a standing wave's "second overtone". A "second overtone" has four nodes and three antinodes. So λ = (2/3)l.

 

[alexfloo]

Thanks a lot! I was erroneously considering the fundamental to be the first overtone.