## College Physics (4th Edition)

The change in frequency heard by the driver is $~-68.8~Hz$
We can convert the car's speed to units of $m/s$: $v = 85~km/h \times \frac{1000~m}{1~km}\times \frac{1~hr}{3600~s} = 23.6~m/s$ We can use the equation for the Doppler effect when the observer is approaching: $f_i = \left(\frac{v+v_o}{v}\right)~f_s$ $f_i = \left(\frac{343~m/s+23.6~m/s}{343~m/s}\right)~(500~Hz)$ $f_i = 534.4~Hz$ We can use the equation for the Doppler effect when the observer is moving away: $f_f = \left(\frac{v-v_o}{v}\right)~f_s$ $f_f = \left(\frac{343~m/s-23.6~m/s}{343~m/s}\right)~(500~Hz)$ $f_f = 465.6~Hz$ We can find the change in frequency: $f_f-f_i = 465.6 - 534.4~Hz = -68.8~Hz$ The change in frequency heard by the driver is $~-68.8~Hz$.