Question

The Mistuned Piano Strings Two identical piano strings of length 0.800 m are each tuned exactly to 480 Hz. The tension in one of the strings is then increased by 1.0%. If they are now struck, what is the beat frequency between the fundamentals of the two strings? SOLUTION Conceptualize As the tension in one of the strings is changed, its fundamental frequency changes. Therefore, when both strings are played, they will have different frequencies and beats be heard. Categorize We must combine our understanding of the waves model for strings with our new knowledge of beats.

Answer 1

Complete Question

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Answer:

The answer is

     $T_2 = 1.008$ % higher than $T_1$

    $T_2 = 0.99$ % lower than $T_1$

Explanation:

   From the question we are told that

         The first string has a frequency of   $f_1 = 230 Hz$

          The period of the beat is  $t_{beat} = 0.99s$

Generally the frequency of the beat is

             $f_{beat} = \frac{1}{t_{beat}}$

  Substituting values

            $f_{beat} = \frac{1}{0.99}$

                   $= 1.01 Hz$

From the question

        $f_2 - f_1 = f_{beat}$   for  $f_2$  having a  higher tension

So

       $f_2 - 230 = 1.01$

               $f_2 = 231.01Hz$

 From the question

            $\frac{f_2}{f_1} = \sqrt{\frac{T_2}{T_1} }$

         $\frac{T_2}{T_1} = \frac{f_2^2}{f_1^2}$

Substituting values

         $\frac{T_2}{T_1} = \frac{(231.01)^2}{(230)^2}$

      $T_2 = 1.008$ % higher than $T_1$

    For $f_2$ having a lower tension

           $f_1 - f_2 = f_{beat}$

  So

       $230 - f_2 = 1.01$

            $f_2 = 230 -1.01$

                  $= 228.99$

  From the question

            $\frac{f_2}{f_1} = \sqrt{\frac{T_2}{T_1} }$

         $\frac{T_2}{T_1} = \frac{f_2^2}{f_1^2}$

    Substituting values

         $\frac{T_2}{T_1} = \frac{(228.99)^2}{(230)^2}$

      $T_2 = 0.99$ % lower than $T_1$