Question

You have a stopped pipe of adjustable length close to a taut 85.0cm, 7.25g wire under a tension of 4170N. 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 (third harmonic). With a very large amplitude. How long should the pipe be?

Answer 1

Answer:

The length is $L_d= 0.069 \ m$

Explanation:

From the question we are told that

               The length of the wire  $L = 85cm = \frac{85}{100} = 0.85m$

                The mass is  $m = 7.25g = \frac{7.25}{1000} = 7.25^10^{-3}kg$

                The tension is  $T = 4170N$

Generally the frequency of  oscillation of a stretched wire is mathematically represented as

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

Where n is the the number of nodes = 3 (i.e the third harmonic)

             $\mu$ is the linear mass density of the wire

 This linear mass density is mathematically represented as

               $\mu = \frac{m}{L}$

Substituting values

            $\mu = \frac{7.2*10^{-3}}{0.85}$

                $= 8.53 *10^{-3} kg/m$

 Substituting values in to the equation for frequency

            $f = \frac{3}{2 80.85} * \sqrt{\frac{4170}{8.53*10^{-3}} }$

               $= 1234Hz$

From the question the we can deduce that the fundamental frequency is equal to the oscillation of a stretched wire

The fundamental frequency is mathematically represented as

        $f = \frac{v}{4L_d}$

Where $L_d$ is the  length of the pipe

          v is the speed of sound with a value of $v = 343m/s$

    Making  $L_d$ the subject of the formula

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

   Substituting values

                   $L_d = \frac{343}{(4)(1234)}$

                        $L_d= 0.069 \ m$

From the question the we can deduce that the fundamental frequency is equal to the oscillation of a stretched wire