Beats Frequency Heard by Moving Observer from Two Sound Sources

To find the number of beats heard per second by observer $O$, we need to calculate the apparent frequencies of the sound received by $O$ from both trains $S_1$ and $S_2$.

1. Conversion of Given Quantities

• Speed of train $S_1$: $v_1 = 108 \text{ km/h} = 108 \times \frac{5}{18} = 30 \text{ m/s}$
• Speed of observer $O$: $v_O = 36 \text{ km/h} = 36 \times \frac{5}{18} = 10 \text{ m/s}$
• Speed of sound in air: $v = 330 \text{ m/s}$
• Frequency emitted by both whistles: $f_0 = 120 \text{ Hz}$

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2. Frequency Received from Stationary Source $S_2$ ($f_2$)

Observer $O$ is moving directly towards the stationary train $S_2$ along the line joining them at speed $v_O = 10 \text{ m/s}$.

Using the Doppler effect formula for a moving observer and a stationary source: $f_2 = f_0 \left( \frac{v + v_O}{v} \right)$

Substituting the values: $f_2 = 120 \left( \frac{330 + 10}{330} \right) = 120 \times \frac{340}{330} = \frac{1360}{11} \text{ Hz} \approx 123.636 \text{ Hz}$

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3. Frequency Received from Moving Source $S_1$ ($f_1$)

From the given geometry:

• $S_2 S_1 = 800 \text{ m}$ (along horizontal axis)
• $S_2 O = 600 \text{ m}$ (along vertical axis)

The direct distance between observer $O$ and train $S_1$ is: $S_1 O = \sqrt{800^2 + 600^2} = 1000 \text{ m}$

Let $\theta$ be the angle between the line of motion of $S_1$ and the line joining $S_1$ to $O$: $\cos\theta = \frac{800}{1000} = 0.8$ $\sin\theta = \frac{600}{1000} = 0.6$

Now, we resolve the velocities along the line joining $S_1$ and $O$:

• Effective velocity of source $S_1$ towards $O$: $v_{s, \text{eff}} = v_1 \cos\theta = 30 \times 0.8 = 24 \text{ m/s}$
• Effective velocity of observer $O$ towards $S_1$: $v_{o, \text{eff}} = v_O \sin\theta = 10 \times 0.6 = 6 \text{ m/s}$

Using the Doppler effect formula for relative motion along the line joining source and observer: $f_1 = f_0 \left( \frac{v + v_{o, \text{eff}}}{v - v_{s, \text{eff}}} \right)$

Substituting the values: $f_1 = 120 \left( \frac{330 + 6}{330 - 24} \right) = 120 \left( \frac{336}{306} \right) = 120 \times \frac{56}{51} = \frac{2240}{17} \text{ Hz} \approx 131.765 \text{ Hz}$

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4. Calculation of Beat Frequency

The number of beats heard per second by observer $O$ is given by the difference between the two apparent frequencies: $\Delta f = |f_1 - f_2| = \frac{2240}{17} - \frac{1360}{11}$

Taking the common denominator ($17 \times 11 = 187$): $\Delta f = \frac{2240 \times 11 - 1360 \times 17}{187} = \frac{24640 - 23120}{187} = \frac{1520}{187} \approx 8.128 \text{ Hz}$

Rounding off to two decimal places, the number of beats heard per second is $8.13$ (or $8.12$ to $8.13$).