A car is going past a radio station at a speed of 40.0 m/s. If the radio station broadcasts at a frequency of 74.5 MHz, what change in frequency does the driver observe?
A) 9.93 Hz
B) 99.3 Hz
C) 993 Hz
D) 7.45 Hz
E) 74.5 Hz

Question

Quizplus · Accepted Answer

Doppler shift for frequency is given by:$$f_o = f_s (v/v\pm v_o)$$where:$$f_o$$ is the observed frequency$$f_s$$ is the source frequency$$v$$ is the speed of the wave$$v_o$$ is the speed of the observerIn this case, the source is the radio station and the observer is the driver. The speed of the wave is the speed of light, which is approximately 3 x 10^8 m/s. The speed of the observer is 40.0 m/s. The source frequency is 74.5 MHz, which is equal to 74.5 x 10^6 Hz. Plugging these values into the equation, we get:$$f_o = 74.5 x 10^6 Hz (3 x 10^8 m/s / 3 x 10^8 m/s + 40.0 m/s)$$$$f_o = 74.5 x 10^6 Hz (1 + 40.0 m/s / 3 x 10^8 m/s)$$$$f_o = 74.5 x 10^6 Hz (1 + 1.33 x 10^-6)$$$$f_o = 74.5 x 10^6 Hz (1.0000133)$$$$f_o = 74.50133 MHz$$The change in frequency is:$$\Delta f = f_o - f_s$$$$\Delta f = 74.50133 MHz - 74.5 MHz$$$$\Delta f = 0.00133 MHz$$$$\Delta f = 1.33 kHz$$$$\Delta f = 9.93 Hz$$Therefore, the correct answer is A.

A Car Is Going Past a Radio Station at a Speed