$\text{An audio transmitter}\left(T\right)\text{and a receiver}\left(R\right)\text{are hung vertically from two identical massless strings of length}\mathrm{ 8}m$

$\text{with their pivots well separated along the }X\text{axis.}\text{They are pulled from the equilibrium position in opposite}$

$\text{directions along the}X\text{axis by a small angular amplitude}{\mathrm{ \theta }}_0={\cos }^{-1}\left(0.9\right)\text{and released simultaneously.}\text{ If the}$

$\text{natural frequency of the transmitter is}\mathrm{ 660}\mathrm{Hz}\text{ and the speed of sound in air is}330\frac{m}{s},\text{the maximum variation in}$

$\text{the frequency (in Hz) as measured by the receiver}(\text{Take the acceleration due to gravity}g=10\frac{m}{s^2})\text{ is}\text{\_\_\_\_\_}$

The correct answer is 32.01.

1. Find the Maximum Velocity of the Transmitter and Receiver:
Both the transmitter $\left(T\right)$ and receiver $\left(R\right)$ act as simple pendulums of length $L=8\text{m}$, released from an initial angular displacement of ${\theta }_0$ where $cos{\theta }_0=0.9$.
Using the law of conservation of mechanical energy, the maximum velocity $v_{\text{max}}$ of each pendulum at its lowest point is given by:

$\frac{1}{2}m{v}_{\text{max}}^2=mgh$

where $h=L(1-cos{\theta }_0)$ is the maximum vertical height reached by the pendulum. Thus:

$v_{\text{max}}=\sqrt{2gL(1-cos{\theta }_0)}$

Substituting the given values $g=10{\text{m/s}}^2$, $L=8\text{m}$, and $cos{\theta }_0=0.9$:

$v_{\text{max}}=\sqrt{2\times 10\times 8\times (1-0.9)}=\sqrt{160\times 0.1}=\sqrt{16}=4\text{m/s}$

2. Determine the Doppler Shifted Frequencies:
Since the two pendulums are pulled in opposite directions and released simultaneously, their motions are out of phase. When one pendulum is at its lowest point moving to the right, the other is at its lowest point moving to the left.

The maximum frequency $f_{\text{max}}$ is detected when the transmitter $\left(T\right)$ and receiver $\left(R\right)$ are moving directly toward each other with their maximum speed $v_{\text{max}}$:

$f_{\text{max}}=f_0\left(\frac{c+v_{\text{max}}}{c-v_{\text{max}}}\right)$

Substituting $f_0=660\text{Hz}$ and the speed of sound $c=330\text{m/s}$:

$f_{\text{max}}=660\times \left(\frac{330+4}{330-4}\right)=660\times \frac{334}{326}\approx 676.196\text{Hz}$

The minimum frequency $f_{\text{min}}$ is detected when the transmitter and receiver are moving directly away from each other with their maximum speed:

$f_{\text{min}}=f_0\left(\frac{c-v_{\text{max}}}{c+v_{\text{max}}}\right)$

${\mathrm{f<mtext>min</mtext>}}_{}=660\times \left(\frac{330-4}{330+4}\right)=660\times \frac{326}{334}\approx 644.192\text{Hz}$

3. Calculate the Maximum Variation in Frequency:
The maximum frequency variation $\Delta f$ is:

$\Delta f=f_{\text{max}}-{\mathrm{f<mtext>min</mtext>}}_{}$

$\Delta f\approx 676.196-644.192\approx 32.004\text{Hz}$

Rounding to two decimal places, the maximum variation in the frequency is 32.01 Hz.