Source moves towards observer at 30 m/s — find apparent frequency

Question

A source of sound of frequency 500 Hz moves towards a stationary observer at 30 m/s. If the speed of sound is 340 m/s, find the apparent frequency heard by the observer.

Solution — Step by Step

Given:

Actual frequency: $f = 500$ Hz
Speed of source: $v_s = 30$ m/s (moving towards observer)
Speed of observer: $v_o = 0$ (stationary)
Speed of sound: $v = 340$ m/s

Doppler formula:

f' = f \cdot \frac{v \pm v_o}{v \mp v_s}

Sign convention: use + in numerator when observer moves toward source; use − in denominator when source moves toward observer.

Observer is stationary: $v_o = 0$ .

Source moves toward observer: use − in denominator (source approaching compresses wavefronts → higher frequency).

f' = f \cdot \frac{v + 0}{v - v_s} = f \cdot \frac{v}{v - v_s}

f' = 500 \times \frac{340}{340 - 30} = 500 \times \frac{340}{310}

f' = 500 \times 1.0968 = \mathbf{548.4 \text{ Hz}}

Rounding: $f' \approx \mathbf{548 \text{ Hz}}$

The apparent frequency (548 Hz) is greater than the actual frequency (500 Hz). This makes physical sense: when the source approaches, wavefronts are compressed (shorter wavelength, higher frequency). The observer hears a higher pitch.

The increase: $548 - 500 = 48$ Hz, roughly a 10% increase — reasonable for a source moving at 30/340 ≈ 9% of the speed of sound.

Why This Works

When a source moves toward an observer, each successive wavefront is emitted slightly closer to the observer. The wavefronts bunch up — the wavelength decreases. Since frequency = speed / wavelength, and the speed of sound is constant (it depends on the medium, not the source), a shorter wavelength means a higher frequency.

The formula captures this: $(v - v_s)$ in the denominator is the effective wavelength spacing (smaller, so frequency is larger). If the source were moving away, we’d use $(v + v_s)$ in the denominator — larger effective spacing, lower frequency.

Alternative Method — Wavelength Approach

Apparent wavelength when source moves toward observer:

\lambda' = \frac{v - v_s}{f} = \frac{310}{500} = 0.62 \text{ m}

Apparent frequency:

f' = \frac{v}{\lambda'} = \frac{340}{0.62} = 548.4 \text{ Hz}

Same answer, different route. This approach clarifies the physics: the wavelength is compressed from $v/f = 340/500 = 0.68$ m to 0.62 m.

Common Mistake

Using $v + v_s$ in the denominator when the source is approaching. This gives a lower frequency than actual — the opposite of what should happen when a source approaches. When source approaches observer: denominator = $v - v_s$ (subtract source velocity). When source recedes: denominator = $v + v_s$ (add source velocity). A memory aid: approaching → frequency increases → denominator must be smaller → subtract.

Question

Solution — Step by Step

Why This Works

Alternative Method — Wavelength Approach

Common Mistake

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