f' = f × (v + v_observer)/(v - v_source). Why the pitch of a passing vehicle changes. Full calculation.

Doppler Effect: Ambulance Moving Towards You at 30 m/s — Apparent Frequency

Question

An ambulance is moving towards a stationary observer at 30 m/s, with its siren blaring at 600 Hz. What is the apparent frequency heard by the observer? (Speed of sound in air = 340 m/s)

Solution — Step by Step

The general Doppler formula is:

f' = f \times \frac{v + v_0}{v - v_s}

Here, $v$ = speed of sound, $v_0$ = speed of observer, $v_s$ = speed of source. The observer is stationary, so $v_0 = 0$ . The source (ambulance) is moving towards the observer, so $v_s = +30$ m/s in the denominator.

This is where most students lose marks. The convention built into this formula is: speeds that decrease the wavelength between source and observer go in the denominator as subtraction, and speeds that increase the effective wavelength go as addition.

Since the ambulance moves towards you, it’s compressing the sound waves — denominator shrinks, frequency goes up. This matches physical intuition.

f' = 600 \times \frac{340 + 0}{340 - 30}

f' = 600 \times \frac{340}{310}

f' = 600 \times 1.0968...

f' = 600 \times \frac{34}{31} \approx 658 \text{ Hz}

The apparent frequency is approximately 658 Hz — noticeably higher than the actual 600 Hz.

Why This Works

When the ambulance moves towards you, it “chases” the sound waves it emitted a moment ago. This bunches up the wavefronts — the wavelength gets shorter. Since $v = f\lambda$ and the wave speed doesn’t change (it’s a property of the medium, not the source), a shorter wavelength means higher frequency.

The moment the ambulance passes you and starts moving away, the opposite happens. Now each successive wavefront is emitted from a position farther from you. The wavelength stretches out, frequency drops. That characteristic weeee-owww pitch drop? That’s the Doppler effect in real time.

This is why the formula has $(v - v_s)$ in the denominator for an approaching source — we’re subtracting the source velocity from the effective wave propagation speed relative to the source-observer gap.

Alternative Method — Using Wavelength Directly

We can solve this by first finding the compressed wavelength.

The source emits 600 waves per second. But in 1 second, it has also moved 30 m closer. So those 600 waves are packed into a distance of:

\lambda' = \frac{v - v_s}{f} = \frac{340 - 30}{600} = \frac{310}{600} \approx 0.517 \text{ m}

Now the observer receives waves at speed 340 m/s with wavelength 0.517 m:

f' = \frac{v}{\lambda'} = \frac{340}{0.517} \approx 658 \text{ Hz}

Same answer. This method is more physical — you can see the wavefront compression happening.

Common Mistake

Swapping $v_s$ and $v_0$ in the formula. A very common error: students put the observer velocity in the denominator and source velocity in the numerator. Always ask yourself — who is moving? Write it out before substituting. The source changes the wavelength (denominator effect), while the observer changes how fast they encounter wavefronts (numerator effect). These are physically different mechanisms.

Memory trick for the signs: “Source chasing → subtract from denominator (frequency rises). Observer chasing → add to numerator (frequency rises).” Both moving towards each other → frequency goes up. Both moving apart → frequency drops. Consistency check before every Doppler problem saves marks.

This question has appeared in CBSE Class 11 board exams as a standard numerical and shows up in JEE Main as a conceptual trap — sometimes they ask what happens after the ambulance passes, expecting you to switch the sign on $v_s$ . The frequency after passing is:

f'_{after} = 600 \times \frac{340}{340 + 30} = 600 \times \frac{340}{370} \approx 551 \text{ Hz}

The swing from 658 Hz to 551 Hz — that’s a 107 Hz drop in perceived pitch, happening in an instant as the vehicle passes.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using Wavelength Directly

Common Mistake

Try These Next