Doppler Effect: Tricky Questions Solved (2)

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Question

A source emits sound of frequency f0=400f_0 = 400 Hz. The source moves toward a stationary wall with speed vs=20v_s = 20 m/s. A stationary observer stands behind the source and hears both the direct sound and the sound reflected by the wall. Find the beat frequency. Speed of sound =340= 340 m/s.

Solution — Step by Step

Source moves away from observer (since source is moving toward wall, away from observer behind it). Observed frequency:

f1=f0vv+vs=400340360=136000360377.78Hzf_1 = f_0 \cdot \frac{v}{v + v_s} = 400 \cdot \frac{340}{360} = \frac{136000}{360} \approx 377.78 \, \text{Hz}

Source moves toward wall:

fwall=f0vvvs=400340320=425Hzf_{wall} = f_0 \cdot \frac{v}{v - v_s} = 400 \cdot \frac{340}{320} = 425 \, \text{Hz}

The wall acts as a stationary source emitting at 425425 Hz.

Observer is stationary, source (the wall) is stationary, so the observer simply hears 425425 Hz directly.

fbeat=freflectedfdirect=425377.78=47.22Hzf_{beat} = |f_{reflected} - f_{direct}| = |425 - 377.78| = 47.22 \, \text{Hz}

Final answer: beat frequency 47.2\approx 47.2 Hz.

Why This Works

Each leg of the journey applies the standard Doppler formula. The wall is treated first as observer (receives fwallf_{wall}), then as source (re-emits at fwallf_{wall}). Between source-wall and wall-observer, frequency does not change again because both parties are stationary in the second leg.

The two simultaneous frequencies at the observer create the beat.

Alternative Method

Some textbooks combine the two formulas into one round-trip Doppler shift:

fref=f0vvvsf_{ref} = f_0 \cdot \frac{v}{v - v_s}

(since wall-to-observer leg has stationary parties, no extra factor). Then beats with f1=f0v/(v+vs)f_1 = f_0 v/(v + v_s).

Students sometimes apply Doppler twice in the wall-to-observer leg too. After reflection, the wall is stationary and the observer is stationary — there is no Doppler shift in this leg. Apply the formula only when something moves.

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