A stationary source at f_s = 440 Hz in air (v = 340 m/s) has a detector moving toward the source at v_d = 10 m/s. What observed frequency f_d is observed?

A stationary source at f_s = 440 Hz in air (v = 340 m/s) has a detector moving toward the source at v_d = 10 m/s. What observed frequency f_d is observed?

• A. 440 Hz
• B. 400 Hz
• C. 454 Hz
• D. 480 Hz

Answer: (C)

The Doppler effect for a sound and a moving observer toward a stationary source increases the observed frequency. The formula to use is f_d = f_s (v + v_d) / v, where v is the speed of sound in the medium, f_s is the source frequency, and v_d is the observer’s speed toward the source. Plugging in the numbers: v = 340 m/s, f_s = 440 Hz, and v_d = 10 m/s gives (v + v_d)/v = (340 + 10)/340 = 350/340 ≈ 1.02941. So f_d ≈ 440 × 1.02941 ≈ 452.94 Hz, which rounds to about 454 Hz. Since the observer moves toward the source, the observed frequency is higher than the source frequency, aligning with the 454 Hz result.

The Doppler effect for a sound and a moving observer toward a stationary source increases the observed frequency. The formula to use is f_d = f_s (v + v_d) / v, where v is the speed of sound in the medium, f_s is the source frequency, and v_d is the observer’s speed toward the source.

Plugging in the numbers: v = 340 m/s, f_s = 440 Hz, and v_d = 10 m/s gives (v + v_d)/v = (340 + 10)/340 = 350/340 ≈ 1.02941. So f_d ≈ 440 × 1.02941 ≈ 452.94 Hz, which rounds to about 454 Hz.

Since the observer moves toward the source, the observed frequency is higher than the source frequency, aligning with the 454 Hz result.