Answer :
Answer:
a) [tex]f_o=454.11Hz[/tex]
b)[tex]f_o=425.89Hz[/tex]
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:
[tex]f_o=\frac{v\pm v_o}{v\pm v_s} *f_s[/tex]
where:
[tex]f_o=Observed\hspace{3}frequency[/tex]
[tex]f_s=Actual\hspace{3}frequency[/tex]
[tex]v=Speed\hspace{3}of\hspace{3}the\hspace{3}sound\hspace{3}waves[/tex]
[tex]v_s=Velocity\hspace{3}of\hspace{3}the\hspace{3}source[/tex]
[tex]v_o=Velocity\hspace{3}of\hspace{3}the\hspace{3}observer[/tex]
Now let's consider the next cases:
[tex]+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[/tex]
[tex]-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[/tex]
[tex]-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[/tex]
[tex]+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[/tex]
The data provided by the problem is:
[tex]f_s=440Hz\\v_o=11m/s[/tex]
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:
[tex]v=343m/s[/tex]
Now, in the first case the observer alone is in motion towards to the source, hence:
[tex]f_o=\frac{v+v_o}{v}*f_s=\frac{343+11}{343} *440=454.1107872Hz[/tex]
Finally, in the second case the observer alone is in motion away from the source, so:
[tex]f_o=\frac{v-v_o}{v}*f_s=\frac{343-11}{343} *425.8892128Hz[/tex]