Respuesta :
Answer: [tex]13.4\ m/s[/tex]
Explanation:
Given
The frequency of the source is [tex]f_o=960\ Hz[/tex]
Change in frequency is [tex]75\ Hz[/tex]
Speed of sound [tex]c=343\ m/s[/tex]
Suppose [tex]v[/tex] is the velocity of the observer
Doppler frequency is given by
[tex]f'=f_o\left(\dfrac{c\pm v_o}{c\pm v_s}\right)[/tex]
Here, the source is at rest
While approaching source, frequency is
[tex]f_1=f_o\left(\dfrac{c+v}{c}\right)\quad \ldots(i)[/tex]
While leaving, frequency is
[tex]f_2=f_o\left(\dfrac{c-v}{c}\right)\quad \ldots(ii)[/tex]
The difference in the frequency is
[tex]\Rightarrow f_1-f_2=75\\\\\Rightarrow f_o\left(\dfrac{c+v}{c}\right)-f_o\left(\dfrac{c-v}{c}\right)=75\\\\\Rightarrow f_o\left(\dfrac{2v}{c}\right)=75\\\\\Rightarrow v=\dfrac{75\times 343}{2\times 960}\\\\\Rightarrow v=13.39\approx 13.4\ m/s[/tex]
