A sound source is fixed at the origin and continuously emits sound...

The position of the source is $S(0, 0) and the listener is at$L(336, -140)$.<br><br>The distance between the source and the listener is$d = \sqrt{336^2 + (-140)^2} = \sqrt{112896 + 19600} = 364 \text{ m}$.<br><br>Assuming the listener moves in the positive$y$-direction, the velocity of the listener is$\vec{v}_L = 26 \hat{j} \text{ m s}^{-1}$.<br><br>The unit vector from the listener to the source is$\hat{u} = \dfrac{-336 \hat{i} + 140 \hat{j}}{364} = -\dfrac{12}{13} \hat{i} + \dfrac{5}{13} \hat{j}$.<br><br>The component of the listener's velocity towards the source is$v_{L \parallel} = \vec{v}_L \cdot \hat{u} = (26 \hat{j}) \cdot \left( -\dfrac{12}{13} \hat{i} + \dfrac{5}{13} \hat{j} \right) = 26 \times \dfrac{5}{13} = 10 \text{ m s}^{-1}$.<br><br>Using the Doppler effect formula, the observed frequency is:<br><br>$f' = f_0 \left(\dfrac{v + v_{L \parallel}}{v} \right)

f' = 660 \left(\dfrac{330 + 10}{330} \right) =$660 \times$\dfrac{340}{330} =$680 \text{ Hz}$.

Answer: $680 \text{ Hz}$