Doppler effect — find apparent frequency when source moves toward stationary observer

Question

A source of sound emitting a frequency of 500 Hz moves toward a stationary observer with a speed of 30 m/s. If the speed of sound in air is 330 m/s, find the apparent frequency heard by the observer.

(JEE Main 2023, similar pattern)

Solution — Step by Step

The source is moving toward the observer, and the observer is stationary. When the source approaches, wavefronts get compressed — the observer encounters more wavefronts per second, so the apparent frequency increases.

We need the standard Doppler formula for sound.

For a source moving toward a stationary observer:

f' = f_0 \left(\frac{v}{v - v_s}\right)

where $f_0$ = original frequency, $v$ = speed of sound, and $v_s$ = speed of source.

Why $v - v_s$ in the denominator? Because the source chases its own wavefronts — the effective wavelength shrinks by the factor $(v - v_s)/f_0$ .

f' = 500 \times \frac{330}{330 - 30} = 500 \times \frac{330}{300}

f' = 500 \times 1.1 = \mathbf{550 \text{ Hz}}

Source moves toward observer, so apparent frequency should be higher than 500 Hz. We got 550 Hz — consistent. If we had obtained a value less than 500, we would know the sign convention was wrong.

Why This Works

The Doppler effect for sound depends on the motion of source and observer relative to the medium (air). When the source moves toward the observer, it effectively reduces the wavelength of the sound waves reaching the observer. Shorter wavelength at the same wave speed means higher frequency.

The key distinction from the light Doppler effect: for sound, it matters who is moving (source vs observer), because sound needs a medium. For light in vacuum, only relative velocity matters.

Sign convention shortcut: if motion reduces the distance between source and observer, use the sign that increases frequency. Source approaching → denominator gets smaller → $f'$ increases. Observer approaching → numerator gets larger → $f'$ increases.

Alternative Method

Use the general Doppler formula directly:

f' = f_0 \left(\frac{v + v_o}{v - v_s}\right)

Here $v_o = 0$ (observer stationary), $v_s = 30$ m/s (source approaching, positive toward observer).

f' = 500 \times \frac{330 + 0}{330 - 30} = 550 \text{ Hz}

Same answer. This general formula handles all cases — just remember the sign convention: velocities toward each other are positive.

Common Mistake

The biggest trap: students swap the sign in the denominator. They write $v + v_s$ instead of $v - v_s$ when the source moves toward the observer. An easy way to avoid this: ask yourself — “should frequency increase or decrease?” Source approaching means higher pitch, so the formula must give $f' > f_0$ . If your formula gives a lower value, your sign is wrong.

Another common error is using this formula for light. The sound Doppler formula does NOT work for electromagnetic waves — light uses the relativistic Doppler formula instead.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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