Question

A train sounds its whistle as it approaches an observer standing at a point near the track. The observer measures a frequency of $216$ Hz as the train approaches and a frequency of $184$ Hz as the train leaves. What is the frequency of its whistle?

• A. $210$ Hz
• B. $190$ Hz
• C. $205$ Hz
• D. $202$ Hz
• E. $200$ Hz

Hint:

In many school-level Doppler problems where $v_s$ is small compared to $v$, the source frequency is roughly the average of the approaching and receding frequencies!

Answer 1

Answer: (E)

Concept: The Doppler Effect describes the change in frequency observed due to the relative motion between a source and an observer. For a stationary observer and moving source:
• $f_{app} = f_s \left( \frac{v}{v - v_s} \right)$ (Approaching)
• $f_{rec} = f_s \left( \frac{v}{v + v_s} \right)$ (Receding)

Step 1: {Set up the equations for the two frequencies.}
Let $f_s$ be the source frequency. $$216 = f_s \left( \frac{v}{v - v_s} \right) \quad \cdots (1)$$ $$184 = f_s \left( \frac{v}{v + v_s} \right) \quad \cdots (2)$$

Step 2: {Relate the two equations.}
From (1), $\frac{v - v_s}{v} = \frac{f_s}{216} \implies 1 - \frac{v_s}{v} = \frac{f_s}{216}$ From (2), $\frac{v + v_s}{v} = \frac{f_s}{184} \implies 1 + \frac{v_s}{v} = \frac{f_s}{184}$

Step 3: {Solve for $f_s$ by adding the equations.}
$$(1 - \frac{v_s}{v}) + (1 + \frac{v_s}{v}) = \frac{f_s}{216} + \frac{f_s}{184}$$ $$2 = f_s \left( \frac{1}{216} + \frac{1}{184} \right)$$ $$2 = f_s \left( \frac{184 + 216}{216 \times 184} \right) = f_s \left( \frac{400}{216 \times 184} \right)$$

Step 4: {Final calculation.}
$$f_s = \frac{2 \times 216 \times 184}{400} = \frac{216 \times 184}{200} = 198.72 \approx 200 \text{ Hz}$$ Alternatively, a common approximation is the harmonic mean or simply $f_s \approx \frac{216 + 184}{2} = 200$ Hz for $v_s \ll v$.