Question

A police car on the side of the road (at rest) uses a radar gun to catch speeders as they approach it. The frequency the radar gun emits is 8 x 109 Hz and the speed limit is 65 mi/hr. What is the difference in frequency of the emitted and returned radar wave

Answer 1

Answer:

$\Delta f = 1.49 \times 10^9 Hz$

Explanation:

Apparent frequency that is received to the speeder is given as

$f_1 = f_0\frac{v + v_o}{v}$

$f_1 = (8 \times 10^9)\frac{340 + v_o}{340}$

here we know that

$v_o = 65 mph = 29 m/s$

now we have

$f_1 = (8 \times 10^9)\frac{340 + 29}{340}$

$f_1 = 8.68 \times 10^9 Hz$

now the frequency that is received back from the speeder is given as

$f_2 = f_1\frac{v}{v- v_o}$

$f_2 = (8.68 \times 10^9}\frac{340}{340 - 29}$

$f_2 = 9.49 \times 10^9 Hz$

So difference is the frequency is given as

$\Delta f = 9.49 \times 10^9 - 8 \times 10^9$

$\Delta f = 1.49 \times 10^9 Hz$