Questions: A Doppler radar sends a pulse at (6.00 times 10^9 ) Hz. It reflects off some clouds, and returns to the radar 672 Hz higher. What is the speed of the clouds? Hint: Remember, reflected waves shift frequencies twice. (Unit = m / s)

Solution Steps

The Doppler effect describes the change in frequency of a wave in relation to an observer moving relative to the source of the wave. For reflected waves, the frequency shift occurs twice: once when the wave moves towards the object and again when it reflects back.

• Initial frequency of the radar pulse: \( f_0 = 6.00 \times 10^9 \, \text{Hz} \)
• Frequency shift: \( \Delta f = 672 \, \text{Hz} \)

The formula for the Doppler effect for a wave reflecting off a moving object is: \[ \Delta f = \frac{2v}{c} f_0 \] where:

• \( \Delta f \) is the change in frequency,
• \( v \) is the speed of the moving object (clouds),
• \( c \) is the speed of light (\( c \approx 3.00 \times 10^8 \, \text{m/s} \)),
• \( f_0 \) is the original frequency of the radar pulse.

Rearrange the formula to solve for \( v \): \[ v = \frac{\Delta f \cdot c}{2 f_0} \]

Substitute the given values into the equation: \[ v = \frac{672 \, \text{Hz} \times 3.00 \times 10^8 \, \text{m/s}}{2 \times 6.00 \times 10^9 \, \text{Hz}} \]

Perform the calculation: \[ v = \frac{672 \times 3.00 \times 10^8}{2 \times 6.00 \times 10^9} \] \[ v = \frac{2.016 \times 10^{11}}{1.20 \times 10^{10}} \] \[ v = 16.8 \, \text{m/s} \]

Final Answer