An audio transmitter (T) and a receiver (R) are hung vertically from two identical massless strings

Q: Why do we use the small angle approximation for $\cos\theta_0$?

Because $\cos\theta_0 = 0.9$ implies $\theta_0$ is around $25^\circ$, which is small enough for the first two terms of the Taylor series ($1 - \theta^2/2$) to give a highly accurate and algebraically clean result without needing a calculator.

Q: Why does the maximum velocity occur at the mean position?

In Simple Harmonic Motion, all the potential energy is converted into kinetic energy at the equilibrium (mean) position. At the extreme ends, the velocity is zero.

Q: Why are both $v'$ terms positive in the numerator and negative in the denominator for $f_{max}$?

When they move towards each other, the observer (receiver) moving towards the source increases the apparent frequency (hence $v + v'$ in the numerator). The source (transmitter) moving towards the observer also increases the frequency (hence $v - v'$ in the denominator).

Q: Can we just approximate $v^2 - v'^2 \approx v^2$ since $v'$ is small?

You could, and it would give $\Delta f \approx 4fvv'/v^2 = 4fv'/v = 4 \times 660 \times 4 / 330 = 32$. In fact, for competitive exams, this binomial approximation is a brilliant time-saver!

Q: What if the pendulums were released at different times?

If they were out of phase, they wouldn't reach their maximum velocities simultaneously. You would have to calculate their instantaneous velocities at the moment they pass each other, which would result in a smaller $\Delta f$.

Question

An audio transmitter (T) and a receiver (R) are hung vertically from two identical massless strings of length 8 m with their pivots well separated along the $X$ axis. They are pulled from the equilibrium position in opposite directions along the $X$ axis by a small angular amplitude $	heta_0 = \cos^{-1} (0.9)$ and released simultaneously. If the natural frequency of the transmitter is 660 Hz and the speed of sound in air is 330 m/s, the maximum variation in the frequency (in Hz) as measured by the receiver (Take the acceleration due to gravity $g = 10 	ext{ m/s}^2$) is ___

Sigma Percentile · Accepted Answer

The problem of the swinging transmitter and receiver is a beautiful symphony of Mechanics and Wave Optics. It tests not just your ability to plug numbers into a formula, but your physical intuition about **Simple Harmonic Motion (SHM)** and the **Doppler Effect**.

### Analyzing the Setup

Imagine the scene: two pendulums, one carrying an audio transmitter and the other a receiver, are pulled in opposite directions and released simultaneously. Because they are identical and released from the same angle $	heta_0$, they will swing in perfect synchrony.

The question asks for the **maximum variation in frequency**. To find this, we must ask ourselves: *When does the Doppler effect cause the greatest shift?* The Doppler shift is maximized when the relative velocity between the source and the observer is at its peak. In SHM, a pendulum reaches its maximum velocity exactly at the lowest point of its swing (the mean position).

Therefore, the extreme frequencies will be heard when both pendulums cross their mean positions. 
1. **Maximum Frequency ($f_{max}$):** When they swing directly towards each other.
2. **Minimum Frequency ($f_{min}$):** When they swing directly away from each other.

### The Master Equation

Before we calculate the frequencies, we need the maximum velocity $v'$ of the pendulums. We are given $\cos	heta_0 = 0.9$. For small angles, the Taylor series expansion is a lifesaver:
$$ \cos	heta_0 \approx 1 - \frac{	heta_0^2}{2} $$
Substituting $0.9$, we find $\frac{	heta_0^2}{2} = 0.1$, which gives $	heta_0 = \frac{1}{\sqrt{5}}$ radians.

The maximum velocity in SHM is given by $v_{max} = A\omega$. Here, the linear amplitude is $A = l	heta_0$ and the angular frequency is $\omega = \sqrt{\frac{g}{l}}$. 
$$ v' = l	heta_0 \sqrt{\frac{g}{l}} = 	heta_0\sqrt{gl} $$
Plugging in $l = 8$ m and $g = 10 	ext{ m/s}^2$, we get $v' = \frac{1}{\sqrt{5}} \sqrt{80} = 4 	ext{ m/s}$.

Now, let's set up the Doppler equations. When moving towards each other:
$$ f_{max} = f \left( \frac{v + v'}{v - v'} ight) $$
When moving away from each other:
$$ f_{min} = f \left( \frac{v - v'}{v + v'} ight) $$

The maximum variation is the difference between these two extremes:
$$ \Delta f = f_{max} - f_{min} = f \left[ \frac{v + v'}{v - v'} - \frac{v - v'}{v + v'} ight] $$
By taking a common denominator and expanding the numerators, the cross terms cancel out beautifully, leaving us with:
$$ \Delta f = f \left[ \frac{4vv'}{v^2 - v'^2} ight] $$

### Final Calculation

Now we substitute our known values: $f = 660 	ext{ Hz}$, $v = 330 	ext{ m/s}$, and $v' = 4 	ext{ m/s}$.
$$ \Delta f = 660 	imes \frac{4 	imes 330 	imes 4}{330^2 - 4^2} $$

Here is where a **smart approximation** separates the masters from the novices. Notice that $330^2$ is $108900$, while $4^2$ is merely $16$. The subtraction of $16$ is practically negligible! We can safely approximate $v^2 - v'^2 \approx v^2$.
$$ \Delta f \approx 660 	imes \frac{4 	imes 330 	imes 4}{330^2} $$
$$ \Delta f \approx 660 	imes \frac{16}{330} = 2 	imes 16 = 32 	ext{ Hz} $$

This elegant approximation not only saves precious time during an exam but also highlights a deep physical truth: when the source velocity is much smaller than the wave speed, the Doppler shift is highly linear. The maximum variation in frequency is exactly **32 Hz**.

Enter Numerical Value:

Visualized Solution

\text{System Setup}

\text{Small Angle Approximation}

\text{Calculating } \theta_0

\text{Maximum Velocity in SHM}

\text{Calculating } v'

\text{Doppler Effect Extremes}

\text{Frequency Formulas}

\text{Maximum Variation } \Delta f

\text{Substituting Values}

\text{Smart Approximation}

\text{The Way Forward}

The Sigma Insight: Doppler Effect

Analyzing the Setup

The Master Equation

Final Calculation

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