Question

Two guitarists attempt to play the same note of wavelength 6.50 cm at the same time, but one of the instruments is slightly out of tune. Consequently, a 17.0-Hz beat frequency is heard between the two instruments. What were the possible wavelengths of the out-of-tune guitar’s note? Express your answers, separated by commas, in centimeters to three significant figures IN cm.

Answer 1

Answer:

The two value of the wavelength for the out of tune guitar is  

$\lambda _2 = (6.48,6.52) \ cm$

Explanation:

From the question we are told that

     The wavelength of the note is $\lambda = 6.50 \ cm = 0.065 \ m$

     The difference in beat frequency is $\Delta f = 17.0 \ Hz$

     

Generally the frequency of the note played by the guitar that is in tune is  

        $f_1 = \frac{v_s}{\lambda}$

Where $v_s$ is the speed of sound with a constant value $v_s = 343 \ m/s$

       $f_1 = \frac{343}{0.0065}$

      $f_1 = 5276.9 \ Hz$

The difference in beat is mathematically represented as

       $\Delta f = |f_1 - f_2|$

Where $f_2$ is the frequency of the sound from the out of tune guitar

     $f_2 =f_1 \pm \Delta f$

substituting values

      $f_2 =f_1 + \Delta f$

      $f_2 = 5276.9 + 17.0$  

     $f_2 = 5293.9 \ Hz$

The wavelength for this frequency is

      $\lambda_2 = \frac{343 }{5293.9}$

     $\lambda_2 = 0.0648 \ m$

    $\lambda_2 = 6.48 \ cm$

For the second value of the second frequency

     $f_2 = f_1 - \Delta f$

     $f_2 = 5276.9 -17$

      $f_2 = 5259.9 Hz$

The wavelength for this frequency is

   $\lambda _2 = \frac{343}{5259.9}$

   $\lambda _2 = 0.0652 \ m$

   $\lambda _2 = 6.52 \ cm$