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(B) Let the speed of the observer be $v_o = v$ and the speed of the source be $v_s = 2v$. At time $t$,the position of the observer is $(0, vt)$ and th

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(B) Let the speed of the observer be $v_o = v$ and the speed of the source be $v_s = 2v$. At time $t$,the position of the observer is $(0, vt)$ and the position of the source is $(2vt, 0)$. The velocity vector of the observer is $\vec{v}_o = v \hat{j}$ and the velocity vector of the source is $\vec{v}_s = 2v \hat{i}$. The unit vector from the source to the observer is $\hat{r} = \frac{-2vt \hat{i} + vt \hat{j}}{\sqrt{(2vt)^2 + (vt)^2}} = \frac{-2 \hat{i} + \hat{j}}{\sqrt{5}}$. The component of the source velocity along the line of sight is $v_{s,r} = \vec{v}_s \cdot \hat{r} = (2v \hat{i}) \cdot \left( \frac{-2 \hat{i} + \hat{j}}{\sqrt{5}} \right) = -\frac{4v}{\sqrt{5}}$. The component of the observer velocity along the line of sight is $v_{o,r} = \vec{v}_o \cdot \hat{r} = (v \hat{j}) \cdot \left( \frac{-2 \hat{i} + \hat{j}}{\sqrt{5}} \right) = \frac{v}{\sqrt{5}}$. The apparent frequency is given by $f = f_0 \left( \frac{c - v_{o,r}}{c - v_{s,r}} \right)$,where $c$ is the speed of sound. Substituting the values,$f = f_0 \left( \frac{c - v/\sqrt{5}}{c + 4v/\sqrt{5}} \right)$. Since $v_{s,r}$ and $v_{o,r}$ are constant,the apparent frequency $f$ is constant over time. Therefore,the graph of $f$ versus $t$ is a horizontal straight line. Comparing this with the given options,Graph $B$ represents a constant frequency $f < f_0$.

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