Question

A friend rides, in turn, the rims of three fast merry-go-rounds while holding a sound source that emits isotropically at a certain frequency. You stand far from each merry-go-round. The frequency you hear for each of your friend's three rides varies as the merry-go-round rotates. The variations in frequency for the three rides are given by the three curves in Fig. 17.5. Rank the curves according to (a) the linear speed of the sound source, (b) the angular speeds of the merry-go-rounds, and (c) the radii r of the merry-go-rounds, greatest first.

Answer 1

Complete Question

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Answer:

a

The Ranking of the curve according to their speed would be equal Rank because    $v_1 =v_2 =v_3$

b

 The first frequency would have a higher rank compared to the other two which will have the same ranking when ranked with respect to their angular velocities because

                                $w_1 >w_2 = w_3$  

c

The ranking of  the second third frequency would be the same but their ranking would be greater than that of the first frequency because

                          $r_2 =r_3 >r_1$

Explanation:

Mathematically Frequency can be represented as

                         $F = \frac{v}{\lambda}$

Where $\lambda$ is the wavelength and v is the velocity

   Now looking at the diagram we see that

          For the  first frequency we have

             Let the wavelength be  $\lambda_1 = \lambda$ , and the frequency  $F_1 = F$

           For  the second frequency

           Let the wavelength be  $\lambda_2 = 2 \lambda$ , and the frequency $F_2 = \frac{F}{2}$

           For  the third frequency

           Let the wavelength be  $\lambda_3 = 2\lambda$ ,  and the frequency $F_3 = \frac{F}{2}$

To obtain v for each of the frequency we make v the subject in the equation above for each frequency

  So,

        For the  first frequency we have

                                 $v_1 = \lambda_1 F_1 = \lambda F$

          For  the second frequency

                               $v_2 = \lambda_2 F_2 = 2 \lambda*\frac{F} {2} = \lambda F$      

           For  the third frequency

                               $v_3 = \lambda_3 F_3 = 2 \lambda*\frac{F} {2} = \lambda F$

Hence

The Ranking of the curve according to their speed would be equal Rank because    $v_1 =v_2 =v_3$

 Mathematically angular speed can be represented as

                           $w = 2 \pi f$

   For the  first frequency we have

                          $w_1 = 2\pi F_1 = 2 \pi F$                        

    For  the second frequency

                        $w_2 = 2 \pi F_2 = 2 \pi \frac{F}{2} = \pi F$

     For  the third frequency

                      $w_3 = 2 \pi F_3 = 2 \pi \frac{F}{2} = \pi F$  

 Hence

          The first frequency would have a higher rank compared to the other two which will have the same ranking when ranked with respect to their angular velocities because

                                $w_1 >w_2 = w_3$  

Mathematically the relationship between the angular velocity and the linear velocity can be represented as

                            $v = wr$

                    =>    $r = \frac{v}{w}$

 Since the linear velocity is constant we have that

                            $r \ \alpha \ \frac{1}{w}$

This means that r varies inversely to the angular velocity ,What this means for ranking due to the radius is that the ranking of  the second third frequency would be the same but their ranking would be greater than that of the first frequency because

                          $r_2 =r_3 >r_1$