Doppler Effect
By The apparent change in the frequency of sound due to relative motion between the source and the observer is given by \begin{align} \nu=\frac{v+u_o}{v-u_s}\nu_0\nonumber \end{align} where, $v$ is the speed of sound in the medium, $u_0$ is the speed of the observer w.r.t. the medium, considered positive when it moves towards the source and negative when it moves away from the source, and $u_s$ is the speed of the source w.r.t. the medium, considered positive when it moves towards the observer and negative when it moves away from the observer.
A Thought Provoking Problem on Doppler Effect
A source S emits sound of frequency $\nu_0$ and an observer O runs towards the source with a speed $u$. The speed of sound in air is $v_0$. According to the equation, the apparent frequency of the sound received by O is \begin{align} \nu^\prime=\nu_0\frac{v_0+u}{v_0}. \end{align}
Now analyze the situation from the observer frame. In this frame, the observer is at rest and the source is moving towards the observer with speed $u$. According to the equation, the apparent frequency of sound received by O must be \begin{align} \nu^{\prime\prime}=\nu_0\frac{v_0}{v_0-u}. \end{align} But $\nu^\prime$ and $\nu^{\prime\prime}$ are different. What is the frequency heard by the observer, $\nu^\prime$ or $\nu^{\prime\prime}$? Where is the fallacy in the above arguments.
Problems from IIT JEE
Problem (IIT JEE 2013): Two vehicles, each moving with speed $u$ on the same horizontal straight road, are approaching each other. Wind blows along the road with velocity $w$. One of these vehicles blows a whistle of frequency $f_1$. An observer in the other vehicle hears the frequency of the whistle to be $f_2$. The speed of sound in still air is $v$. The correct statement(s) is (are),
- If the wind blows from the observer to the source, $f_2 > f_1$.
- If the wind blows from the source to the observer, $f_2 > f_1$.
- If the wind blows from the observer to the source, $f_2 < f_1$.
- If the wind blows from the source to the observer, $f_2 < f_1$.
Solution: Doppler's effect equation is, \begin{align} \label{yvb:eqn:1} f_2=\left[\frac{v+u_o}{v-u_s}\right]f_1. \end{align} Let the wind speed be $w$ and wind moves towards the source in case (i) and towards the observer in case (ii) (see figure).
In the case (i), $u_o$, the speed of observer w.r.t. medium considered positive when it moves towards the source is $+(u-w)$ and $u_s$, the speed of source w.r.t. medium considered positive when it moves towards observer is $+(u+w)$. Substitute these values in first equation to get, \begin{align} f_2=\left[\frac{v+(u-w)}{v-(u+w)}\right]f_1, \nonumber \end{align} which is greater than $f_1$ (assuming $u>w$). In the case (ii), $u_o$ is $+(u+w)$ and $u_s$ is $+(u-w)$. Substitute these values in first equation to get, \begin{align} f_2=\left[\frac{v+(u+w)}{v-(u-w)}\right]f_1, \nonumber \end{align} which is again greater than $f_1$. The readers are encouraged to show that first equation can be generalized to, \begin{align} f_2=\left[\frac{v+w+u_o}{v+w-u_s}\right]f_1, \nonumber \end{align} where $w$ is positive if the wind blows in the direction of sound and negative if it blows opposite to the direction of sound. Thus, A and B are correct options.
