Question

A street musician sounds the A string of his violin, producing a tone of 440 Hz, What frequency does a bicyclist hear as he a) approaches and b) recedes from the musician with a speed of 11.0 m/s?

Answer 1

Answer:

a) $f_o=454.11Hz$

b)$f_o=425.89Hz$

Explanation:

Let´s use Doppler effect, in order to calculate the observed frequency by the byciclist. The Doppler effect equation for a general case is given by:

$f_o=\frac{v\pm v_o}{v\pm v_s} *f_s$

where:

$f_o=Observed\hspace{3}frequency$

$f_s=Actual\hspace{3}frequency$

$v=Speed\hspace{3}of\hspace{3}the\hspace{3}sound\hspace{3}waves$

$v_s=Velocity\hspace{3}of\hspace{3}the\hspace{3}source$

$v_o=Velocity\hspace{3}of\hspace{3}the\hspace{3}observer$

Now let's consider the next cases:

$+v_o\hspace{3}is\hspace{3}used\hspace{3}when\hspace{3}the\hspace{3}observer\hspace{3}moves\hspace{3}towards\hspace{3}the\hspace{3}source$

$-v_o\hspace{3}is\hspace{3}used\hspace{3}when\hspace{3}the\hspace{3}observer\hspace{3}moves\hspace{3}away\hspace{3}from\hspace{3}the\hspace{3}source$

$-v_s\hspace{3}is\hspace{3}used\hspace{3}when\hspace{3}the\hspace{3}source\hspace{3}moves\hspace{3}towards\hspace{3}the\hspace{3}observer$

$+v_s\hspace{3}is\hspace{3}used\hspace{3}when\hspace{3}the\hspace{3}source\hspace{3}moves\hspace{3}away\hspace{3}from\hspace{3}the\hspace{3}observer$

The data provided by the problem is:

$f_s=440Hz\\v_o=11m/s$

The problem don't give us aditional information about the medium, so let's assume the medium is the air, so the speed of sound in air is:

$v=343m/s$

Now, in the first case the observer alone is in motion towards to the source, hence:

$f_o=\frac{v+v_o}{v}*f_s=\frac{343+11}{343} *440=454.1107872Hz$

Finally, in the second case the observer alone is in motion away from the source, so:

$f_o=\frac{v-v_o}{v}*f_s=\frac{343-11}{343} *425.8892128Hz$