A TV station has a weather radar which is monitoring an approaching thunderstorm

Question

A TV station has a weather radar which is monitoring an approaching thunderstorm. The radar emits radio waves at a frequency of 2900 MHz . The thunderstorm line is approaching the radar installation at 45.0 km/h

Radio waves are reflected from the water drops in the thunderstorms, and the reflected waves are detected back at the installation. As measured by the receiver at the installation, is the frequency of these reflected waves greater or less than 2900 MHz ?

By what amount does the frequency of the reflected waves differ from 2900 MHz ?

Correct A TV station has a weather radar which is monitoring an approaching thunderstorm. The radar emits radio waves at a frequency of 2900 MHz. The thunderstorm line is approaching the radar installation at 45.0 km/h Part C Radio waves are reflected from the water drops in the thunderstorms, and the reflected waves are detected back at the installation. As measured by the receiver at the installation, is the frequency of these reflected waves greater or less than 2900 MHz? greater less Submit My Answers Give Up Correct Part D By what amount does the frequency of the reflected waves differ from 2900 MHz? ? as= 121 Hz Submit My Answers Give Up

Explanation / Answer

Velocity of thunderstorm, vt = 45 km/h = 45*5/18 = 12.5 m/s

initial frequency, f = 2900 MHz = 2.9*109 Hz

speed of radio wave, v = 3*108 m/s

for 1 st stage source (radar) is stationary and observer ( thunderstorm) is moving

frequncy observed by thunderstorm, f' = f (v - vt) /v

f' = 2.9*109 ( 3*108 + 12.5) / 3*108 = 2900000120.83 Hz

for 2nd stage source (thunderstorm) is moving and observer (radar) is stationary.

frequncy observed by thunderstorm, f'' = f' v  /(v - vt)

f'' = 2.9*109 ( 3*108 + 12.5) * 3*108/ (3*108 * (3*108 - 12.5)) = 2900000241.67 Hz

So change in frequency = 241.67 Hz

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