Solution to 2008 Problem 45

The wavelength that the person observes \lambda is related to the wavelength that the siren emits \lambda_0 by
\begin{align*}\lambda = \lambda_0 \left(\frac{v\pm v_S}{v \mp v_O} \right)\end{align*}
where v_S is the velocity of the siren (the source) relative to the medium, v_O is the velocity of the person (the observer) relative to the medium, and v is the velocity of sound in the medium. Here, v_S = v_0 = 55 \mbox{ m/s} and v = 330 \mbox{ m/s}. Relative to the medium, the siren is moving away from the person, and the person is moving towards the siren, so we take the plus sign in both the numerator and the denominator. Thus,
\begin{align*}\lambda = \lambda_0 \left(\frac{v + v_S}{v + v_O} \right) = \lambda_0 \left(\frac{330 \mbox{ m/s} + 55 \mbox{ m...
or in terms of frequencies
\begin{align*}f = f_0\end{align*}
f_0 = 1200 \mbox{ Hz}, so
\begin{align*}f = \boxed{1200 \mbox{ Hz}}\end{align*}
as well. Therefore, answer (C) is correct.

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