Animated Solution for Physics - Waves: Comprehension Passage

Welcome, future engineers and physicists! Today, we are going to dive into a truly elegant problem from the JEE Advanced 2023 paper. This question beautifully marries the concepts of kinematics in two dimensions with the Doppler Effect and Beat Frequencies. It is not just a test of your memory of formulas, but a profound test of your ability to visualize physical situations.

Imagine you are standing at point P, the detector, at the bottom of a massive circular track. You are blindfolded, relying entirely on your ears. Somewhere on this track, a siren (Source $S_2$) is racing around anti-clockwise, and another siren (Source $S_1$) is sitting right at the center. Both are blaring at exactly $656\text{ Hz}$.

Let's break down the symphony of physics happening here!

First, let's map out our coordinates. Let the center of the circle be the origin $O(0,0)$. The radius of the circle is $R$. The detector is at point $P$, which is at the bottom, so its coordinates are $(0,-R)$. Point $R$ is diametrically opposite to $P$, so it sits at the top at $(0,R)$. Point $Q$ is equidistant from $P$ and $R$, which places it perfectly on the right side of the horizontal axis at $(R,0)$.

The speed of sound in air is given as $C=324\text{ m/s}$. The moving source $S_2$ has a uniform speed of $v=4\sqrt{2}\text{ m/s}$.

The Doppler Shift from a Tangential Source

In the first scenario, only $S_2$ is emitting sound, and it is currently passing through point $Q$. What frequency do you hear at $P$?

To solve any Doppler effect problem, the golden rule is: Only the component of velocity along the line of sight matters.

At point $Q(R,0)$, $S_2$ is moving anti-clockwise. This means its velocity vector is purely tangential, pointing straight up along the positive y-axis: $\vec{v}=v\hat{j}$.

Now, draw a line from the source $Q$ to the detector $P$. This is our line of sight. The vector from $Q$ to $P$ is $-R\hat{i}-R\hat{j}$. If we look at the geometry, the triangle $OPQ$ is a right-angled isosceles triangle. The line $QP$ makes a $45^{\circ }$ angle with the negative y-axis.

Since the velocity is pointing straight up (positive y-axis), the angle between the velocity vector and the line of sight (pointing down-left) is $135^{\circ }$. This means the source is effectively moving away from the detector. The component of velocity directed away from the detector is:

Now, we bring in the master equation for the Doppler effect when a source is moving away from a stationary observer:

Substitute our known values:

Notice how beautifully the numbers are crafted for you. $328\times 2=656$.

So, when $S_2$ is at $Q$, the detector hears a lower frequency of $648\text{ Hz}$.

The Zero Doppler Shift and Approaching Source

Now, the plot thickens. Both sources are turned on. $S_2$ has traveled to point $R$ at the top of the circle, and $S_1$ (which was at the center) starts moving directly towards the detector $P$ at $4\text{ m/s}$. We need to find the beat frequency.

Let's analyze $S_2$ first. At point $R(0,R)$, it is moving anti-clockwise, so its velocity is purely horizontal, pointing to the left: $\vec{v}=-v\hat{i}$. The line of sight from $R$ to $P$ is straight down along the y-axis. What is the angle between the velocity vector (horizontal) and the line of sight (vertical)? It is exactly $90^{\circ }$!

Since $\cos (90^{\circ })=0$, the component of velocity along the line of sight is zero. There is absolutely no Doppler shift! The frequency heard from $S_2$ is its true frequency:

Now, let's look at $S_1$. It is moving directly towards the detector with a speed of $v_{S1}=4\text{ m/s}$. Because it is approaching, the sound waves get compressed, and the apparent frequency increases. We use the Doppler formula with a minus sign in the denominator:

Substitute the values:

Let's simplify the fraction. Dividing by $4$, we get $\frac{81}{80}$.

The detector at $P$ is now receiving two distinct frequencies simultaneously: $656\text{ Hz}$ from $S_2$ and $664.2\text{ Hz}$ from $S_1$. When two sound waves of slightly different frequencies superimpose, they create a phenomenon called beats—a periodic fluctuation in volume.

The beat frequency is simply the absolute difference between the two frequencies:

And there we have it! By carefully breaking down the velocity vectors and applying the Doppler principles step-by-step, we have conquered this beautiful JEE Advanced problem. Always remember, physics is not just math; it is the geometry of reality. Keep visualizing, and keep practicing!