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.

Solution to 2008 Problem 45