Beat Frequency Observed from Two Moving Sound Sources

To find the beat frequency heard by the stationary observer, we use the Doppler effect for sound.

1. Geometric Setup

Let the horizontal line of motion of the two walking men be along the $x$-axis ($y = 0$), and the observer $O$ be positioned at $(0, 12\text{ m})$ in the $xy$-plane.

The two men are walking in the same direction ($+x$-direction):

• The person in front is at position $x_1$ with speed $v_1 = 1.0\text{ m s}^{-1}$.
• The person behind is at position $x_2$ with speed $v_2 = 2.0\text{ m s}^{-1}$.

At the given instant, the two men are separated by $10\text{ m}$ and are equidistant from the observer. Thus: $x_1 = +5\text{ m}, \quad x_2 = -5\text{ m}$

The distance of each person from the observer is: $r = \sqrt{x^2 + y^2} = \sqrt{(\pm 5)^2 + 12^2} = \sqrt{25 + 144} = 13\text{ m}$

The angle $\theta$ that the line of motion makes with the line joining each person to the observer satisfies: $\cos\theta = \frac{5}{13}$

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2. Components of Velocity along the Line of Sight

• For the person behind (Source 2): This person is at $x_2 = -5\text{ m}$ and moving in the $+x$-direction, which is towards the observer. The component of velocity directed towards the observer is: $v_{2\parallel} = v_2 \cos\theta = 2.0 \times \frac{5}{13} = \frac{10}{13}\text{ m s}^{-1}$

• For the person in front (Source 1): This person is at $x_1 = +5\text{ m}$ and moving in the $+x$-direction, which is away from the observer. The component of velocity directed away from the observer is: $v_{1\parallel} = v_1 \cos\theta = 1.0 \times \frac{5}{13} = \frac{5}{13}\text{ m s}^{-1}$

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3. Frequencies Perceived by the Stationary Observer

The speed of sound in air is $v = 330\text{ m s}^{-1}$, and the emitted frequency is $f_0 = 1430\text{ Hz}$.

Using the Doppler effect formula for a moving source and a stationary observer:

1. Frequency perceived from the person behind: $f_2 = f_0 \left( \frac{v}{v - v_{2\parallel}} \right) = 1430 \left( \frac{330}{330 - \frac{10}{13}} \right) = 1430 \left( \frac{4290}{4290 - 10} \right) = 1430 \times \frac{4290}{4280}\text{ Hz}$

2. Frequency perceived from the person in front: $f_1 = f_0 \left( \frac{v}{v + v_{1\parallel}} \right) = 1430 \left( \frac{330}{330 + \frac{5}{13}} \right) = 1430 \left( \frac{4290}{4290 + 5} \right) = 1430 \times \frac{4290}{4295}\text{ Hz}$

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4. Beat Frequency

The beat frequency $\Delta f = f_2 - f_1$ is given by: $\Delta f = 1430 \times 4290 \left( \frac{1}{4280} - \frac{1}{4295} \right) = 1430 \times 4290 \times \frac{15}{4280 \times 4295}$

Since $v_{1\parallel}, v_{2\parallel} \ll v$, we can also use the binomial approximation: $f_2 \approx f_0 \left(1 + \frac{v_{2\parallel}}{v}\right)$ $f_1 \approx f_0 \left(1 - \frac{v_{1\parallel}}{v}\right)$

Thus: $\Delta f \approx f_0 \left( \frac{v_{2\parallel} + v_{1\parallel}}{v} \right) = 1430 \left( \frac{\frac{10}{13} + \frac{5}{13}}{330} \right) = 1430 \times \frac{15}{13 \times 330}$

Simplifying: $\frac{1430}{13} = 110$ $\Delta f = 110 \times \frac{15}{330} = 110 \times \frac{1}{22} = \frac{110}{22} = 5.00\text{ Hz}$

Calculating with the exact expression: $\Delta f = 1433.34 - 1428.34 = 5.00\text{ Hz}$

The beat frequency heard by the stationary observer is 5.00 (or 5).