a radio wave of frequency 840 MHz is sent towards an aeroplane. the frequency of the radio echo is 2.8 KHz more than the original frequency. the velocity of aero plane..

To determine the velocity of the airplane based on the frequency shift of the radio wave, we can apply the Doppler effect, which describes how the frequency of a wave changes for an observer moving relative to the source of the wave. In this case, the radio wave is sent towards the airplane, and the echo received has a higher frequency due to the airplane's motion.

Understanding the Doppler Effect

The Doppler effect occurs when there is relative motion between a wave source and an observer. If the source is moving towards the observer, the observed frequency increases, while if it is moving away, the observed frequency decreases. The formula for the observed frequency (\(f'\)) when the source is moving towards the observer is given by:

f' = f (1 + v/c)

Where:

• f' = observed frequency
• f = original frequency
• v = velocity of the observer (airplane)
• c = speed of light (approximately \(3 \times 10^8\) m/s)

Given Values

In this scenario:

• Original frequency, \(f = 840 \text{ MHz} = 840 \times 10^6 \text{ Hz}\)
• Frequency shift, \(\Delta f = 2.8 \text{ KHz} = 2.8 \times 10^3 \text{ Hz}\)

The observed frequency can be calculated as:

f' = f + \Delta f = 840 \times 10^6 \text{ Hz} + 2.8 \times 10^3 \text{ Hz}

f' = 840,002,800 \text{ Hz}

Calculating the Velocity of the Airplane

Now, we can substitute the values into the Doppler effect formula to find the velocity of the airplane:

f' = f (1 + v/c)

Rearranging this gives us:

1 + v/c = f'/f

v/c = (f'/f) - 1

v = c \left( \frac{f'}{f} - 1 \right)

Substituting the known values:

v = 3 \times 10^8 \left( \frac{840,002,800}{840,000,000} - 1 \right)

Calculating the fraction:

v = 3 \times 10^8 \left( \frac{840,002,800}{840,000,000} - 1 \right) = 3 \times 10^8 \left( 1.000003333 - 1 \right)

v = 3 \times 10^8 \times 0.000003333

v ≈ 1000 \text{ m/s}

Final Thoughts

The calculated velocity of the airplane is approximately 1000 m/s. This example illustrates how the Doppler effect can be used to determine the speed of an object based on the change in frequency of waves it reflects. In practical applications, this principle is vital in radar technology and various fields of physics and engineering.