Question

A child has an ear canal that is 1.3 cm long. Assume the speed of sound is v = 344 m/s.

Answer 1

Answer:

The  frequencies are $(f, f_1) = (6615.4 \ Hz , 19846.2\ Hz)$

Explanation:

From the question we are told that

  The  length of the ear canal is  $l = 1.3 \ cm =\frac{1.3}{100} = 0.013 \ m$

   The  speed of sound is assumed to be  $v_s = 344 \ m/s$

Now  taking look at a typical  ear canal  we see that we assume it is  a  closed pipe

   Now the fundamental harmonics for the pipe(ear canal) is mathematically represented as

            $f = \frac{v_s}{4 * l }$

 substituting values  

          $f = \frac{344}{4 * 0.013 }$

         $f = 6615.4 \ Hz$

Also the the second harmonic for the pipe (ear canal) is mathematically represented as

        $f_1 = \frac{3v_s}{4 * l}$

 substituting values  

       $f_1 = \frac{3 * 344}{4 * 0.013}$

       $f_1 = 19846.2 \ Hz$

Given that sound would be loudest in the pipe at the frequency, it implies that the child  will have an increased audible sensitivity at this  frequencies