Question

You have a stopped pipe of adjustable length close to a taut 62.0-cm, 7.25-g wire under a tension of 3910 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? (Express your answer with the appropriate units.)

Answer 1

Answer:

L = 6 cm

Explanation:

Second overtone of the wire is same as third harmonic

so its frequency is given as

$f = \frac{3}{2L}\sqrt{\frac{T}{\mu}}$

here we know that

$\mu = \frac{M}{L}$

$\mu = \frac{7.25 \times 10^{-3}}{0.62}$

$\mu = 0.0117 kg/m$

now we have

$f = \frac{3}{2\times 0.62}\sqrt{\frac{3910}{0.0117}}$

$f = 1398.6 Hz$

Now fundamental frequency of sound in a pipe is given as

$f = \frac{v}{4L}$

$1398.6 = \frac{340}{4L}$

$L = 0.06 m$

$L = 6 cm$