Two cars $$X$$ and $$Y$$ are approaching each other with velocities 36 km h$$^{-1}$$ and 72 km h$$^{-1}$$ respectively. The frequency of a whistle sound as emitted by a passenger in car $$X$$, heard by the passenger in car $$Y$$ is 1320 Hz. If the velocity of sound in air is 340 ms$$^{-1}$$, the actual frequency of the whistle sound produced is _________ Hz.

We need to find the actual frequency of the whistle sound produced by a passenger in car $$X$$, given the observed frequency heard by a passenger in car $$Y$$ as they approach each other.

1. Convert Velocities to SI Units ($$\text{m s}^{-1}$$)

The velocities of the cars are given in $$\text{km h}^{-1}$$. We convert them to meters per second ($$\text{m s}^{-1}$$) by multiplying by $$\frac{5}{18}$$:

• Velocity of Source car $$X$$ ($$v_s$$):

  $$v_s = 36 \times \frac{5}{18} = 10\text{ m s}^{-1}$$

• Velocity of Observer car $$Y$$ ($$v_o$$):

  $$v_o = 72 \times \frac{5}{18} = 20\text{ m s}^{-1}$$

2. Apply the Doppler Effect Formula

When the source and the observer are approaching each other, the apparent frequency ($$f'$$) heard by the observer increases and is given by the formula:

$$f' = f \left( \frac{v + v_o}{v - v_s} \right)$$

Where:

• $$f' = 1320\text{ Hz}$$ (Observed frequency)
• $$f$$ = Actual frequency of the whistle (to be calculated)
• $$v = 340\text{ m s}^{-1}$$ (Velocity of sound in air)
• $$v_o = 20\text{ m s}^{-1}$$ (Velocity of the observer)
• $$v_s = 10\text{ m s}^{-1}$$ (Velocity of the source)

3. Substitute the Values and Solve for $f$

Substitute the known quantities into the Doppler Effect equation:

$$1320 = f \left( \frac{340 + 20}{340 - 10} \right)$$

$$1320 = f \left( \frac{360}{330} \right)$$

Simplify the fraction inside the parentheses:

$$\frac{360}{330} = \frac{12}{11}$$

$$1320 = f \left( \frac{12}{11} \right)$$

Now, isolate the actual frequency ($$f$$):

$$f = 1320 \times \frac{11}{12}$$

$$f = 110 \times 11 = 1210\text{ Hz}$$