Question

You have a stopped pipe of adjustable length close to a taut 62.0 cm, 7.25 g wire under a tension of 3310 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. The speed of sound in air is 344 m/s. How long should the pipe be?

Answer 1

Answer:

L' = 6.7 cm

Explanation:

given,

length of pipe = 62 cm = 0.62 m

mass of wire = 7.25 g = 0.00725 kg

Tension = 3310 N

speed of sound = 344 m/s

the second overtone frequency

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

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

$\mu = \dfrac{0.00725}{0.62}$

$\mu = 0.01169 kg/m$

$f = \dfrac{3}{2 (0.62)}\sqrt{\dfrac{3310}{0.01169}}$

f = 1,287.37 Hz

fundamental frequency

$f = \dfrac{v}{4L'}$

the length of pipe

$L'= \dfrac{v}{4f}$

$L'= \dfrac{343}{4\times 1287.37}$

L' = 0.067 m

L' = 6.7 cm