Question

You have a stopped pipe of adjustable length close to a taut 62.0 cm, 7.25 g wire under a tension of 4310 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:

the length of the pipe should be 11.7 cm

Explanation:

given information:

the length of the taut, L = 62 cm = 0.62 m

wire's mass = 7.25 g = 7.25 x 10⁻³ kg

tension, F = 4310 N

speed sound, $v_{s}$ = 344 m/s

to find the length of the pipe, we first calculate the speed of the wave on a string by:

v = √F/μ                                    (1)

where

v = speed

F = tension

μ = linear density

μ = m/L, m is mass, and L is the length

thus,

v = √FL/m

  = √(4310)(0.62)/(7.25 x 10⁻³)

  = 607.11 m/s

now we find the frequency of  the wire using

$f = n\frac{v}{2L}$ (n=1,2,3,.....)                     (2)

where

f = frequency

second overtone with very large amplitude, n = 3

so,

$f = 3\frac{607.11}{2(0.62)}$

  = 1468.81

now we can calculate the length of the pipe using the second equation. the frequency of the pipe and the wire is the same, therefore

$f = n\frac{v}{2L}$

n = 1, fundamental frequency

$L = \frac{v}{2f}$

   = $\frac{344}{2 (1468.81)}$

   = 0.117 m

   = 11.7 cm