Question

11. A tight guitar string has a frequency of 540 Hz as its third harmonic. What will be its fundamental frequency if it is fingered at a length of only 70% of its original length

Answer 1

Answer:

The frequency is  $f_n = 257.1 \ Hz$

 

Explanation:

From the question we are told that

    The third harmonic frequency of the tight guitar string is  $f_3 = 540 \ Hz$

     

Let the original length be  L  

   Then the length at which it is fingered is  0.7 L

Generally the fundamental  is mathematically represented as

         $f = \frac{v_s}{ 2L}$

Now when it finger at 70% it original length is

      $f_n = \frac{v}{2 * (0.7 L)}$

      $f_n = \frac{v}{1.4 L}$

Here v  the velocity of sound

  So  

         $\frac{f_n}{f} = \frac{\frac{v}{1.4L} }{\frac{v}{2L} }$

Also the fundamental frequency for the original length can also be represented as

       $f = \frac{f_3}{3}$

substituting values

          $f = \frac{540}{3}$

          $f = 180 \ Hz$

So

       $\frac{f_n}{180} = \frac{\frac{v}{1.4L} }{\frac{v}{2L} }$

=>  $f_n =\frac{180}{0.7}$

=>   $f_n = 257.1 \ Hz$