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?

Answers

  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

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