Question

Baseball scouts often use a radar gun to measure the speed of a pitch. One particular model of radar gun emits a microwave signal at a frequency of 10.525 GHz. What will be the increase in frequency if these waves are reflected from a 95.0mi/h fastball headed straight toward the gun?

Answer 1

Since the Units presented are not in the International System we will proceed to convert them. We know that,

$1 mi/h = 0.447 m/s$

So the speed in SI would be

$V=95mi/h(\frac{0.447m/s}{1mi/h})$

$V=42.465 m/s$

The change in frequency when the wave is reflected is

$f'=f(1+\frac{V}{c})$

Or we can rearrange the equation as

$f' = f + f\frac{V}{c}$

f' = Apparent frequency

f = Original Frequency

c = Speed of light

$f'-f = f\frac{V}{c}$

$\Delta f = f\frac{V}{c}$

Replacing,

$\Delta f = (10.525*10^9)(\frac{42.465}{3*10^8})$

$\Delta f =1489.8 Hz$

Since the waves are reflected, hence the change in frequency at the gun is equal to twice the change in frequency

$\Delta f_T = 2 \Delta f$

$\Delta f_T = 2(1489.8Hz)$

$\Delta f_T = 2979.63Hz$

Therefore the increase in frequency is 2979.63Hz