Chapter 39, Problem 39.69AP

A Doppler weather radar station broadcasts a pulse of radio waves at frequency 2.85 GHz. From a relatively small batch of raindrops at bearing 38.6° east of north, the station receives a reflected pulse after 180 µs with a frequency shifted upward by 254 Hz. From a similar batch of raindrops at bearing 39.6° east of north, the station receives a reflected pulse after the same time delay, with a frequency shifted downward by 254 Hz. These pulses have the highest and lowest frequencies the station receives, (a) Calculate the radial velocity components of both batches of raindrops. (b) Assume that these raindrops are swirling in a uniformly rotating vortex. Find the angular speed of their rotation.

To determine

The radial velocity components of both batches of raindrops.

Answer to Problem 39.69AP

The radial velocity component of first batch of raindrops is $13.4\text{m}/\text{s}$ toward the station and the radial velocity component of second batch of raindrops is $13.4\text{m}/\text{s}$ away from the station.

Explanation of Solution

Given info: The frequency of radio waves is $2.85\text{GHz}$, bearing of raindrops is $38.6°$ east of north, the time after which station receives pulse is $180\text{μs}$, the frequency shifts upward by a value equal to $254\text{Hz}$, the bearing of raindrops from a similar batch is $39.6°$ east of north, the time after which the station receives pulse is $180\text{μs}$ and the frequency shifts downward by a value equal to $254\text{Hz}$.

Write the equation of frequency of radio waves received.

$f\text{'}=f\sqrt{\frac{c+v}{c-v}}$

Here,

$f\text{'}$ is the frequency of radio waves received which is reflected by the raindrops.

$f$ is the frequency of radio waves.

$c$ is the speed of light.

$v$ is the speed of the raindrops.

Write the equation of frequency of radio waves received for another upward shift.

$f_1\text{'}=f\text{'}\sqrt{\frac{c+v}{c-v}}$

Here,

$f_1\text{'}$ is the frequency of radio waves received for another upward shift.

Substitute $f\sqrt{\frac{c+v}{c-v}}$ for $f\text{'}$ in above equation.

$\begin{array}{c}{f}_1\text{'}=\left(f\sqrt{\frac{c+v}{c-v}}\right)\sqrt{\frac{c+v}{c-v}}\\ f_1\text{'}=f\left(\frac{c+v}{c-v}\right)\\ v=c\left(\frac{f_1\text{'}-f}{f_1\text{'}+f}\right)\end{array}$ (1)

The frequency of radio waves received for upward shift is,

$f_1\text{'}=f+f_1$

Here,

$f_{\text{1}}$ is the frequency shifted upward.

Substitute $2.85\text{GHz}$ for $f$ and $254\text{Hz}$ for $f_1$ in above equation to find $f_1\text{'}$.

$\begin{array}{c}{f}_1\text{'}=2.85\text{GHz}\left(\frac{{10}^9\text{Hz}}{1\text{GHz}}\right)+254\text{Hz}\\ =\text{2,850,000,254}\text{Hz}\end{array}$

Thus, the value of $f_1\text{'}$ is $\text{2,850,000,254}\text{Hz}$.

Substitute $3.0\times {10}^8\text{m}/\text{s}$ for $c$, $\text{2,850,000,254}\text{Hz}$ for $f_1\text{'}$ and $2.85\text{GHz}$ for $f$ in equation (1) to find $v$.

$\begin{array}{c}v=\left(3.0\times {10}^8\text{m}/\text{s}\right)\left[\frac{\left(\text{2,850,000,254}\text{Hz}\right)-\left\{2.85\text{GHz}\left(\frac{{10}^9\text{Hz}}{1\text{GHz}}\right)\right\}}{\left(\text{2,850,000,254}\text{Hz}\right)+\left\{2.85\text{GHz}\left(\frac{{10}^9\text{Hz}}{1\text{GHz}}\right)\right\}}\right]\\ =13.368\text{m}/\text{s}\\ \approx 13.4\text{m}/\text{s}\end{array}$

Thus, the radial velocity of the radio waves with the upward shift is $13.4\text{m}/\text{s}$ towards the station.

The frequency of radio waves received for downward shift is,

$f_1\text{'}=f+f_1$

Here,

$f_{\text{1}}$ is the frequency shifted downward.

Substitute $2.85\text{GHz}$ for $f$ and $-254\text{Hz}$ for $f_1$ in above equation to find $f_1\text{'}$.

$\begin{array}{c}{f}_1\text{'}=2.85\text{GHz}\left(\frac{{10}^9\text{Hz}}{1\text{GHz}}\right)-254\text{Hz}\\ =\text{2,849,999,746}\text{Hz}\end{array}$

Thus, the value of $f_1\text{'}$ is $\text{2,849,999,746}\text{Hz}$.

Substitute $3.0\times {10}^8\text{m}/\text{s}$ for $c$, $\text{2,849,999,746}\text{Hz}$ for $f_1\text{'}$ and $2.85\text{GHz}$ for $f$ in equation (1) to find $v$.

$\begin{array}{c}v=\left(3.0\times {10}^8\text{m}/\text{s}\right)\left[\frac{\left(\text{2,849,999,746}\text{Hz}\right)-\left\{2.85\text{GHz}\left(\frac{{10}^9\text{Hz}}{1\text{GHz}}\right)\right\}}{\left(\text{2,849,999,746}\text{Hz}\right)+\left\{2.85\text{GHz}\left(\frac{{10}^9\text{Hz}}{1\text{GHz}}\right)\right\}}\right]\\ =-13.368\text{m}/\text{s}\\ \approx -13.4\text{m}/\text{s}\end{array}$

The radial velocity of the radio waves with the downward shift is $13.4\text{m}/\text{s}$ away from the station.

Conclusion:

Therefore, the radial velocity component of first batch of raindrops is $13.4\text{m}/\text{s}$ toward the station and the radial velocity component of second batch of raindrops is $13.4\text{m}/\text{s}$ away from the station.

To determine

The angular speed of the rotation.

Answer to Problem 39.69AP

The angular speed of the rotation is $0.0567\text{rad}/\text{s}$.

Explanation of Solution

Given info: The frequency of radio waves is $2.85\text{GHz}$, bearing of raindrops is $38.6°$ east of north, the time after which station receives pulse is $180\text{μs}$, the frequency shifts upward by a value equal to $254\text{Hz}$, the bearing of raindrops from a similar batch is $39.6°$ east of north, the time after which the station receives pulse is $180\text{μs}$ and the frequency shifts downward by a value equal to $254\text{Hz}$.

Write the equation of angular speed of the vortex.

$\omega =\frac{v}{r}$ (2)

Here,

$\omega$ is the angular speed of the vortex.

$r$ is the radius of vortex.

Write the equation for radius of vortex.

$r=\frac{1}{2}x\theta$

Here,

$x$ is the one way distance covered by the rain.

$\theta$ is the angle made by raindrops.

The radio waves travels towards the rain and comes back.

Write the equation for one way distance covered by the rain.

$x=\frac{1}{2}c\left(\Delta t\right)$

Here,

$\Delta t$ is the time after which the station receives pulse.

Substitute $\frac{1}{2}c\left(\Delta t\right)$ for $x$ in above equation to find $r$.

$\begin{array}{l}r=\frac{1}{2}\left(\frac{1}{2}c\left(\Delta t\right)\right)\theta \\ r=\frac{1}{4}c\Delta t\theta \end{array}$

Substitute $\frac{1}{4}c\left(\Delta t\right)\theta$ for $r$ in equation (2) to find $\omega$.

$\begin{array}{c}\omega =\frac{v}{\left(\frac{1}{4}c\left(\Delta t\right)\theta \right)}\\ \omega =\frac{4v}{c\Delta t\theta }\end{array}$

Substitute $13.4\text{m}/\text{s}$ for $v$, $3.0\times {10}^8\text{m}/\text{s}$ for $c$, $180\text{μs}$ for $\Delta t$ and $1°$ for $\theta$ in above equation to find $\omega$.

$\begin{array}{c}\omega =\frac{4\left(13.4\text{m}/\text{s}\right)}{\left(3.0\times {10}^8\text{m}/\text{s}\right)\left\{180\text{μs}\left(\frac{1\text{s}}{{10}^6\text{μs}}\right)\right\}\left\{1°\left(\frac{\pi \text{rad}}{180°}\right)\right\}}\\ =0.0567\text{rad}/\text{s}\end{array}$

Conclusion:

Therefore, the angular speed of the rotation is $0.0567\text{rad}/\text{s}$.