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Question

(A) When a source is stationary and a reflector (car) approaches it with speed $v$,the frequency $f'$ received by the car is $f' = f_0 \frac{c+v}{c}$.

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(A) When a source is stationary and a reflector (car) approaches it with speed $v$,the frequency $f'$ received by the car is $f' = f_0 \frac{c+v}{c}$.This frequency is then reflected back to the source. The car acts as a moving source,so the frequency $f''$ heard by the stationary source is $f'' = f' \frac{c}{c-v} = f_0 \frac{c+v}{c-v}$.Since $v \ll c$,we use the binomial approximation: $f'' = f_0 (1 + v/c)(1 - v/c)^{-1} \approx f_0 (1 + v/c)(1 + v/c) \approx f_0 (1 + 2v/c)$.The frequency shift for a single car is $\Delta f = f'' - f_0 = f_0 (2v/c)$.For two cars with speeds $v_1$ and $v_2$,the difference in reflected frequencies is $\Delta f_{diff} = |f''_1 - f''_2| = f_0 \frac{2(v_1 - v_2)}{c}$.Given $\Delta f_{diff} = 0.012 f_0$,we have $0.012 f_0 = f_0 \frac{2 \Delta v}{c}$.$\Delta v = 0.006 	imes c = 0.006 	imes 330 \ m/s = 1.98 \ m/s$.Converting to $km/h$: $\Delta v = 1.98 	imes 3.6 \ km/h \approx 7.128 \ km/h$.The nearest integer is $7 \ km/h$.

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