Doppler effect calculations

The relationship

$$f' = f\left(\frac{v_w}{v_w \pm v_s}\right)$$

• $f'$ — the observed frequency (Hz)

• $f$ — the frequency emitted by the source (Hz)

• $v_w$ — the speed of the wave in the medium (m s⁻¹) — for sound in air, usually $340$ m s⁻¹

• $v_s$ — the speed of the source (m s⁻¹)

• This is the form on the resource sheet, and it applies to a stationary observer with a moving source — the only case in this standard.

Choosing the sign

The $\pm$ in the denominator is the whole difficulty of this page, and there is a reliable way to get it right:

• Source approaching → use the minus sign → smaller denominator → $f' > f$ (higher pitch).

• Source receding → use the plus sign → larger denominator → $f' < f$ (lower pitch).

• The memory hook: approaching subtracts.

• The check that never fails: decide first whether the answer should be higher or lower, then confirm your calculated $f'$ agrees. If you expected a higher pitch and got a lower number, you used the wrong sign.

Setting the calculation out

• Write down $f$, $v_w$ and $v_s$ with their units.
• State whether the source is approaching or receding, and therefore which sign you are using.
• Predict whether $f'$ will be larger or smaller than $f$.
• Substitute and evaluate the denominator first.
• Check the answer against your prediction.

Related quantities

• The observed wavelength follows from the wave equation using the unchanged wave speed:

$$\lambda' = \frac{v_w}{f'}$$

• The change in frequency is $\Delta f = |f' - f|$, and questions sometimes ask for this rather than $f'$ itself.
• If a question gives $f'$ and asks for the source speed, rearrange:

$$v_s = v_w\left|\frac{f' - f}{f'}\right|$$

• but it is usually safer to substitute the numbers into the original relationship and solve, rather than memorising a rearranged form.

The approach and recede pair

• A common question gives both the approaching and receding frequencies and asks for the source speed or the emitted frequency.
• The two observed frequencies are not symmetric about the emitted frequency: the rise on approach is always larger than the drop on recession.
  • For a source at $30$ m s⁻¹ emitting $500$ Hz: approaching gives $548$ Hz (up $48$), receding gives $459$ Hz (down $41$).
• The reason is the algebra: dividing by $(v_w - v_s)$ amplifies more than dividing by $(v_w + v_s)$ reduces. Stating this asymmetry is a strong Excellence observation.