1. Understanding the Physical Setup

The problem involves two identical pendulums, one acting as an audio transmitter (source) and the other as a receiver. They are released simultaneously from the same angular amplitude but in opposite directions. This means they will oscillate in Simple Harmonic Motion (SHM) with a phase difference of $\pi$. The frequency of the sound heard by the receiver will change due to the Doppler effect because of the relative motion between the source and the receiver.

2. Calculating the Maximum Speed of the Pendulum Bobs

The maximum speed of each pendulum bob (transmitter and receiver) occurs at the mean (lowest) position of its swing. We can find this speed using the principle of conservation of mechanical energy.

Let $L$ be the length of the string and $\theta_0$ be the maximum angular amplitude. The maximum height of the bob from the mean position is $h_{max} = L - L\cos\theta_0 = L(1 - \cos\theta_0)$.

At the extreme position, the bob has zero kinetic energy and maximum potential energy, $PE_{max} = mgh_{max}$. At the mean position, the potential energy is zero (taking it as the reference level), and the kinetic energy is maximum, $KE_{max} = \frac{1}{2}mv_{max}^2$.

By conservation of energy: $KE_{max} = PE_{max}$ $\frac{1}{2}mv_{max}^2 = mgL(1 - \cos\theta_0)$ $v_{max} = \sqrt{2gL(1 - \cos\theta_0)}$

Given values are:

• Length of the string, $L = 8$ m
• Acceleration due to gravity, $g = 10 \mathrm{~m/s^2}$
• $\theta_0 = \cos^{-1}(0.9)$, which means $\cos\theta_0 = 0.9$

Substituting these values: $v_{max} = \sqrt{2 \times 10 \times 8 \times (1 - 0.9)}$ $v_{max} = \sqrt{160 \times 0.1}$ $v_{max} = \sqrt{16} = 4 \mathrm{~m/s}$ So, the maximum speed of both the transmitter ($v_S$) and the receiver ($v_R$) is 4 m/s.

3. Applying the Doppler Effect

The apparent frequency $f'$ heard by a receiver is given by the Doppler effect formula: $f' = f_0 \left( \frac{v \pm v_R}{v \mp v_S} \right)$ where:

• $f_0$ is the natural frequency of the source = 660 Hz.
• $v$ is the speed of sound in air = 330 m/s.
• $v_R$ is the speed of the receiver.
• $v_S$ is the speed of the source. The signs are chosen based on the direction of motion: top signs for approach, bottom signs for separation.

Maximum Frequency ($f'_{max}$): The frequency heard will be maximum when the source and receiver are moving towards each other with their maximum speeds. Since they oscillate out of phase, this occurs when both are passing through their equilibrium positions. $v_S = v_{max} = 4 \mathrm{~m/s}$ $v_R = v_{max} = 4 \mathrm{~m/s}$ $f'_{max} = f_0 \left( \frac{v + v_R}{v - v_S} \right) = 660 \left( \frac{330 + 4}{330 - 4} \right) = 660 \left( \frac{334}{326} \right) \mathrm{~Hz}$

Minimum Frequency ($f'_{min}$): The frequency heard will be minimum when the source and receiver are moving away from each other with their maximum speeds. This also occurs when both are passing through their equilibrium positions. $f'_{min} = f_0 \left( \frac{v - v_R}{v + v_S} \right) = 660 \left( \frac{330 - 4}{330 + 4} \right) = 660 \left( \frac{326}{334} \right) \mathrm{~Hz}$

4. Calculating the Maximum Variation in Frequency

The maximum variation in frequency is the difference between the maximum and minimum observed frequencies: $\Delta f = f'_{max} - f'_{min}$ $\Delta f = 660 \left( \frac{334}{326} - \frac{326}{334} \right)$ $\Delta f = 660 \left( \frac{334^2 - 326^2}{326 \times 334} \right)$ Using the difference of squares formula, $a^2 - b^2 = (a-b)(a+b)$: $\Delta f = 660 \left( \frac{(334 - 326)(334 + 326)}{326 \times 334} \right)$ $\Delta f = 660 \left( \frac{8 \times 660}{326 \times 334} \right) = \frac{660 \times 8 \times 660}{108884} \approx 32.0048 \mathrm{~Hz}$

Alternatively, since $v_S, v_R \ll v$, we can use the approximation for relative motion. The maximum relative speed of approach is $v_{rel} = v_S + v_R = 4 + 4 = 8$ m/s. The maximum variation in frequency is approximately: $\Delta f \approx f_0 \left( \frac{2 v_{rel}}{v} \right) = 660 \left( \frac{2 \times (v_S+v_R)}{v} \right) = 660 \left( \frac{2 \times 8}{330} \right)$ $\Delta f = \frac{660 \times 16}{330} = 2 \times 16 = 32 \mathrm{~Hz}$

The exact calculation gives a value very close to 32. As an integer answer is required, we take 32.

Final Answer

The maximum variation in the frequency is 32 Hz.