Chapter 12, Problem 82GP

To determine

The speed of car.

Answer to Problem 82GP

  $$ The speed of the car is $114.3\text{m/s}$ .

Explanation of Solution

Given:

The sound of a car is dropped by a full octave.

Formula used:

Write the expression for apparent frequency.

  $\mathrm{f}\text{'}=\frac{\mathrm{v}}{\mathrm{v}-{\mathrm{v}}_{\mathrm{c}\mathrm{a}\mathrm{r}}}\mathrm{f}$

  ${\mathrm{v}}_{\mathrm{c}\mathrm{a}\mathrm{r}}$ is the velocity of the train.

  $\mathrm{v}$ is the speed of the sound.

  $\mathrm{v}=343\text{m/s}$

In the first case the source is moving towards the listener & in the second case the source is moving away from the listener.

  $\begin{array}{l}\mathrm{c}\mathrm{a}\mathrm{s}\mathrm{e}1:\\ \mathrm{f}\text{'}=\frac{343}{343-{\mathrm{v}}_{\text{car}}}\mathrm{f}\mathrm{...........................}(1)\\ \mathrm{c}\mathrm{a}\mathrm{s}\mathrm{e}2:\\ \mathrm{f}\text{'}\text{'}=\frac{343}{343+{\mathrm{v}}_{\text{car}}}\mathrm{f}\mathrm{...........................}(2)\end{array}$

As the sound of the car is dropped by a full octave.

  $\mathrm{f}\text{'}=2\mathrm{f}\text{'}\text{'}\mathrm{.........................................................}(3)$

Calculation:

From equation (1) ,(2) & (3):

  $\begin{array}{l}\frac{343}{343-{\mathrm{v}}_{\text{car}}}=2\left(\frac{343}{343+{\mathrm{v}}_{\text{car}}}\right)\\ 343+{\mathrm{v}}_{\text{car}}=686-2{\mathrm{v}}_{\text{car}}\\ 3{\mathrm{v}}_{\text{car}}=343\\ {\mathrm{v}}_{\text{car}}=\frac{343}{3}=114.3\text{m/s}\end{array}$

  $\begin{array}{l}\frac{538}{486}=\frac{343+{\mathrm{v}}_{\mathrm{s}}}{343-{\mathrm{v}}_{\mathrm{s}}}\\ 1.107(343-{\mathrm{v}}_{\mathrm{s}})=343+{\mathrm{v}}_{\mathrm{s}}\\ 379.7-1.107{\mathrm{v}}_{\mathrm{s}}=343+{\mathrm{v}}_{\mathrm{s}}\\ 36.7=2.107{\mathrm{v}}_{\mathrm{s}}\\ {\mathrm{v}}_{\mathrm{s}}=\frac{36.7}{2.107}=17.42\text{m/s}\end{array}$

Conclusion:

The velocity of the train is $114.3\text{m/s}$ .

  $\begin{array}{l}\frac{538}{486}=\frac{343+{\mathrm{v}}_{\mathrm{s}}}{343-{\mathrm{v}}_{\mathrm{s}}}\\ 1.107(343-{\mathrm{v}}_{\mathrm{s}})=343+{\mathrm{v}}_{\mathrm{s}}\\ 379.7-1.107{\mathrm{v}}_{\mathrm{s}}=343+{\mathrm{v}}_{\mathrm{s}}\\ 36.7=2.107{\mathrm{v}}_{\mathrm{s}}\\ {\mathrm{v}}_{\mathrm{s}}=\frac{36.7}{2.107}=17.42\mathrm{m}/\mathrm{s}\end{array}$

Conclusion:

The velocity of the train is $17.4\mathrm{m}/\mathrm{s}$ .