Chapter 17, Problem 17.71CP

(a)

To determine

The frequency heard by the passengers in the car.

Answer to Problem 17.71CP

The frequency heard by the passengers in the car is $531\text{Hz}$.

Explanation of Solution

Given info: The speed of the train is $25.0\text{m}/\text{s}$, the distance of the car is $30.0\text{m}$ and the frequency of the horn is $500\text{Hz}$

Consider the following figure.

Figure (1)

In right angle triangle $\text{OST}$,

$OS=\sqrt{a^2+b^2}$

Substitute $30\text{m}$ for $a$ and $40\text{m}$ for $b$ in above expression.

$\begin{array}{c}OS=\sqrt{{\left(30\text{m}\right)}^2+40{\text{m}}^2}\\ =50\text{m}\end{array}$

The value of the $OS$ is $50\text{m}$.

In triangle $\text{OST}$,

$\cos {\theta }_{\text{s}}=\frac{TS}{OS}$

Substitute $50\text{m}$ for $OS$ and $40\text{m}$ for $TS$ in above expression.

$\begin{array}{c}\cos {\theta }_{\text{s}}=\frac{40\text{m}}{50\text{m}}\\ =\frac{4}{5}\end{array}$

The value of the $\cos {\theta }_{\text{s}}$ is $\frac{4}{5}$.

The expression for the frequency heard by the passengers in the car is,

$f\text{'}=\left(\frac{v+v_o\cos {\theta }_{\text{o}}}{v-v_{\text{s}}\cos {\theta }_{\text{s}}}\right)f$

Here,

$v_{\text{o}}$ is the speed of the observer.

$v_{\text{s}}$ is the speed of the source.

$f$ is the original frequency.

Substitute $25.0\text{m}/\text{s}$ for $v_{\text{s}}$, $0\text{m}/\text{s}$ for $v_{\text{o}}$, $\frac{4}{5}$ for $\cos {\theta }_{\text{s}}$, $500\text{Hz}$ for $f$ and $343\text{m}/\text{s}$ for $v$ in above expression.

$\begin{array}{c}f\text{'}=\left(\frac{343\text{m}/\text{s}+0\text{m}/\text{s}\times \cos {\theta }_{\text{o}}}{343\text{m}/\text{s}-25.0\text{m}/\text{s}\times \frac{4}{5}}\right)\times 500\text{Hz}\\ =\text{530}\text{.95}\text{Hz}\\ \approx \text{531}\text{Hz}\end{array}$

Conclusion:

Therefore the frequency heard by the passengers in the car is $531\text{Hz}$.

(b)

To determine

The range of frequencies heard by the passenger in the car.

Answer to Problem 17.71CP

The range of frequencies heard by the passenger in the car is $466\text{Hz}\text{to}\text{539}\text{Hz}$.

Explanation of Solution

Given info: The speed of the train is $25.0\text{m}/\text{s}$, the distance of the car is $30.0\text{m}$ and the frequency of the horn is $500\text{Hz}$.

Since the observer and source are moving away from each other so the value of the angles becomes equal to zero.

The expression for the frequency heard by the passengers in the car is,

$f\text{'}=\left(\frac{v+v_o\cos {\theta }_{\text{o}}}{v-v_{\text{s}}\cos {\theta }_{\text{s}}}\right)f$ (1)

For the case when the train is arrived:

Substitute $25.0\text{m}/\text{s}$ for $v_{\text{s}}$, $0\text{m}/\text{s}$ for $v_{\text{o}}$, $1$ for $\cos {\theta }_{\text{s}}$, $500\text{Hz}$ for $f$ and $343\text{m}/\text{s}$ for $v$ in above equation (1).

$\begin{array}{c}f\text{'}=\left(\frac{343\text{m}/\text{s}+0\text{m}/\text{s}\times \cos {\theta }_{\text{o}}}{343\text{m}/\text{s}-25.0\text{m}/\text{s}\times 1}\right)\times 500\text{Hz}\\ =\text{539}\text{Hz}\end{array}$

For the case when train is arriving:

Substitute $-25.0\text{m}/\text{s}$ for $v_{\text{s}}$, $0\text{m}/\text{s}$ for $v_{\text{o}}$, $1$ for $\cos {\theta }_{\text{s}}$, $500\text{Hz}$ for $f$ and $343\text{m}/\text{s}$ for $v$ in above equation (1).

$\begin{array}{c}f\text{'}\text{'}=\left(\frac{343\text{m}/\text{s}+0\text{m}/\text{s}\times \cos {\theta }_{\text{o}}}{343\text{m}/\text{s}+25.0\text{m}/\text{s}\times 1}\right)\times 500\text{Hz}\\ =466\text{Hz}\end{array}$

Conclusion:

Therefore the range of frequencies heard by the passenger in the car is $466\text{Hz}\text{to}\text{539}\text{Hz}$.

(c)

To determine

The frequency heard by the passengers in the car.

Answer to Problem 17.71CP

The frequency heard by the passengers in the car is $551.44\text{Hz}$.

Explanation of Solution

Given info: The speed of the train is $25.0\text{m}/\text{s}$, the distance of the car is $30.0\text{m}$ and the frequency of the horn is $500\text{Hz}$.

The expression for the frequency heard by the passengers in the car is,

$f\text{'}=\left(\frac{v+v_o\cos {\theta }_{\text{o}}}{v-v_{\text{s}}\cos {\theta }_{\text{s}}}\right)f$

Substitute $40.0\text{m}/\text{s}$ for $v_{\text{s}}$, $0\text{m}/\text{s}$ for $v_{\text{o}}$, $\frac{4}{5}$ for $\cos {\theta }_{\text{s}}$, $500\text{Hz}$ for $f$ and $343\text{m}/\text{s}$ for $v$ in above expression.

$\begin{array}{c}f\text{'}=\left(\frac{343\text{m}/\text{s}+0\text{m}/\text{s}\times \cos {\theta }_{\text{o}}}{343\text{m}/\text{s}-40.0\text{m}/\text{s}\times \frac{4}{5}}\right)\times 500\text{Hz}\\ =\text{551}\text{.44}\text{Hz}\end{array}$

Conclusion:

Therefore the frequency heard by the passengers in the car is $551.44\text{Hz}$.