Chapter 39, Problem 69AP

(a)

To determine

The radial velocity components of both batches of raindrops.

Answer to Problem 69AP

The radial velocity components of first batch of raindrops is $\underset{\_}{13.4\text{m/s}}$ towards the station, and $\underset{\_}{13.4\text{m/s}}$ away from the station.

Explanation of Solution

The first batch of raindrops are moving towards the radio station.

Here there is relative motion between the raindrop and the station. So apply Doppler’s equation to the situation.

Write the expression for the enhanced frequency of radio waves received by the first batch of raindrops.

  ${f^{\prime }}_1=f\sqrt{\frac{c+v}{c-v}}$                                                                                                               (I)

Here, ${f^{\prime }}_1$ is the Doppler enhanced frequency received by the first batch of raindrops, $f$ is the original frequency of radio waves transmitted from radio station, $c$ is the speed of light, and $v$ is the speed of radio waves.

The radio waves transmitted towards the first batch of raindrop gets reflected towards the radio station. There is an upward Doppler shift in the frequency of the reflected wave.

Write the expression for the enhanced frequency of the reflected radio waves from first batch of rain drops.

  $f_1{}^{\prime \text{}\prime }={f^{\prime }}_1\sqrt{\frac{c+v}{c-v}}$                                                                                                           (II)

Here, $f_1{}^{\prime \text{}\prime }$ is the enhanced frequency of the reflected radio waves from the first batch of rain drops.

Use expression (I) in (II).

  $\begin{array}{c}{f}_1{}^{\prime \text{}\prime }=f\sqrt{\frac{c+v}{c-v}}\sqrt{\frac{c+v}{c-v}}\\ =f\left(\frac{c+v}{c-v}\right)\end{array}$                                                                                                 (III)

Due to the relative motion of rain drops and the pulse, $f$ suffers a upward Doppler shift from $f$ to ${f^{″}}_1$.

Write the expression to find $f_1{}^{\prime \text{}\prime }$.

  $f_1{}^{\prime \text{}\prime }=f+\Delta f$                                                                                                               (IV)

Here, $\Delta f$ is the upward Doppler shift in frequency.

Use expression (IV) in (III).

  $f+\Delta f=f\left(\frac{c+v}{c-v}\right)$                                                                                                     (V)

Solve expression (V) for $\left(\frac{c+v}{c-v}\right)$.

  $\begin{array}{c}\left(\frac{c+v}{c-v}\right)=\frac{f+\Delta f}{f}\\ =1+\frac{\Delta f}{f}\end{array}$                                                                                                       (VI)

Simplify expression (VI) to find $v$.

    $\begin{array}{c}\left(c+v\right)f=\left(f+\Delta f\right)\left(c-v\right)\\ cf+vf=cf-vf+\Delta fc-\Delta fv\\ 2vf=\Delta fc-\Delta fv\end{array}$                                                                                (VII)

Take terms containing $v$ to left side and solve for $v$.

    $\begin{array}{l}2vf+\Delta fv=\Delta fc\\ v\left(2f+\Delta f\right)=\Delta fc\\ v=\frac{\Delta fc}{\left(2f+\Delta f\right)}\end{array}$                                                                                                  (VIII)

Similarly repeat the calculations for the second batch of raindrops also. Here the frequency is enhanced in down ward direction. So replace $+\Delta f$ by $-\Delta f$ in equation (IV).

Conclusion:

Substitute $2.85\text{GHz}$ for $f$, $254\text{Hz}$ for $\Delta f$, and $3.00\times {10}^8\text{m/s}$ for $c$ in equation (VIII) and solve for $v$ of first batch of rain drops.

    $\begin{array}{c}v=\frac{\left(254\text{Hz}\right)\left(3.00\times {10}^8\text{m/s}\right)}{2\left(2.85\text{GHz}\times \frac{1\text{Hz}}{{10}^{-9}\text{GHz}}\right)+\left(254\text{Hz}\right)}\\ =13.4\text{m/s}\end{array}$

Substitute $2.85\text{GHz}$ for $f$, $-254\text{Hz}$ for $\Delta f$, and $3.00\times {10}^8\text{m/s}$ for $c$ in equation (VIII) and solve for $v$ of second batch of raindrops.

  $\begin{array}{c}v=\frac{\left(-254\text{Hz}\right)\left(3.00\times {10}^8\text{m/s}\right)}{2\left(2.85\text{GHz}\times \frac{1\text{Hz}}{{10}^{-9}\text{GHz}}\right)-\left(254\text{Hz}\right)}\\ =-13.4\text{m/s}\end{array}$

Therefore, the radial velocity components of first batch of raindrops is $\underset{\_}{13.4\text{m/s}}$ towards the station, and $\underset{\_}{13.4\text{m/s}}$ away from the station.

(b)

To determine

Angular speed of rotation of rotation of the rain drops.

Answer to Problem 69AP

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

Explanation of Solution

The first batch of rain drops is at bearing of $38.6°$ and the second batch of raindrop is at bearing $39.6°$. Therefore the batch of the raindrop whirls about a vortex at $1°$ of arc.

The time taken by the radio wave to travel from station to rain and come back is $180\text{μs}$.

Write the expression for one way distance travelled by the radio waves.

  $r=\frac{1}{2}ct$                                                                                                                    (VII)

Here, $r$ is the one way distance travelled by the radio waves, $c$ is the speed of radio waves, and $t$ is the time taken.

Write the expression for the diameter of the vortex where the rain drops are whirling.

  $s=r\theta$                                                                                                                   (VIII)

Here, $s$ is the diameter of the vortex, $r$ is the one way distance travelled by radio waves, and $\theta$ is the angle of whirling of rain drops.

Write the expression for the angular sped of rotation of rain drop about the vortex in terms of the diameter.

  $\begin{array}{c}\omega =\frac{v}{\frac{s}{2}}\\ =\frac{2v}{s}\end{array}$                                                                                                                      (IX)

Here, $\omega$ is the angular speed, and $v$ is the radial speed of rain drops.

Conclusion:

Substitute $3.00\times {10}^8\text{m/s}$ for $c$, and $180\text{μs}$ for $t$ in equation (VII) to find $d$.

  $\begin{array}{c}d=\frac{1}{2}\left(3.00\times {10}^8\text{m/s}\right)\left(180\text{μs}\times \frac{1\text{s}}{{10}^6\text{μs}}\right)\\ =27000\text{m}\end{array}$

Substitute $27000\text{m}$ for $r$, and $1°$ for $\theta$ in equation (VIII) to find $s$.

  $\begin{array}{c}s=\left(27000\text{m}\right)\left(1°\right)\left(\frac{\pi \text{rad}}{180°}\right)\\ =471\text{m}\end{array}$

Substitute $471\text{m}$ for $s$ , and $13.4\text{m/s}$ for $v$ in equation (IX) to find $\omega$.

  $\begin{array}{c}\omega =\frac{2\left(13.4\text{m/s}\right)}{471\text{m}}\\ =0.0567\text{rad/s}\end{array}$

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