Question

A standing man is watching an oncoming firetruck that emits a 900-Hz sound from its siren. If its velocity is 30.5 m/s, and the speed of sound is 345 m/s on this day,

(a) What frequency does the person receive?

(b) What frequency does she receive after the firetruck passed?

Answer 1

(a) We can find the frequency received by the person from the Doppler Shift formula:

$$f_o=f_s(\dfrac{v}{v-v_s}),$$ $$f_o=900\ Hz\cdot(\dfrac{345\ \dfrac{m}{s}}{345\ \dfrac{m}{s}-30.5\ \dfrac{m}{s}})=987\ Hz.$$

(b) We can find the frequency received by the person after the firetruck passed from the Doppler Shift formula:

$$f_o=f_s(\dfrac{v}{v+v_s}),$$ $$f_o=900\ Hz\cdot(\dfrac{345\ \dfrac{m}{s}}{345\ \dfrac{m}{s}+30.5\ \dfrac{m}{s}})=827\ Hz.$$