Chapter 15, Problem 48A

(a)

To determine

To calculate: The frequency perceived by a detector on a stationary train.

Answer to Problem 48A

The frequency of sound perceived by a detector on a stationary stationis $\text{280 Hz}$ .

Explanation of Solution

Given:

Speed of sound in air ( $v$ ) = $+343\text{ m/s}$

Velocity of train (source) ( $v_s$ ) = $\text{31}\text{.0 m/s}$

Velocity of detector train ( $v_d$ ) = $0\text{ m/s}$ (at rest)

Frequency of wave ( $f_s$ ) = $\text{305 Hz}$

The train is moving away from the detector.

Formula used:

Doppler Effect equation:

  $f_d=f_s\left(\frac{v-v_d}{v-v_s}\right)$

Where $f_d$ represents frequency perceived by the detector, $f_s$ represents wave’s frequency, $v$ represents speed of sound in air, $v_s$ represents velocity of source and $v_d$ represents velocity of detector.

Calculation:

As per problem,

The train is moving away from the detector, so $v_s=-31.0\text{ m/s}$

Now put the value of $f_s=305\text{ Hz}$ , $v=343\text{ m/s}$ , $v_d=0\text{ m/s}$ and $v_s=-31.0\text{ m/s}$ in the formula

  $\begin{array}{l}{f}_d=f_s\left(\frac{v-v_d}{v-v_s}\right)\\ f_d=305\text{ Hz}\left(\frac{343\text{ m/s}-0\text{ m/s}}{343\text{ m/s}-(-31.0\text{ m/s})}\right)\\ f_d=305\text{ Hz}\left(\frac{343\text{ m/s}}{374\text{ m/s}}\right)\\ f_d=280\text{ Hz}\end{array}$

Conclusion:

Thus, the frequency perceived by the detectoris $280\text{ Hz}$.

(b)

To determine

To calculate:The frequency perceived by a train moving away from the source train.

Answer to Problem 48A

The frequency perceived by the train moving away from the source train is $297\text{ Hz}$.

Explanation of Solution

Given:

Speed of sound in air ( $v$ ) = $+343\text{ m/s}$

Velocity of train (source) ( $v_s$ ) = $\text{31}\text{.0 m/s}$

Velocity of detector train ( $v_d$ ) = $21.0\text{ m/s}$

Frequency of wave ( $f_s$ ) = $\text{305 Hz}$

Both the trains are moving in opposite direction from each other.

Formula used:

Doppler Effect equation:

  $f_d=f_s\left(\frac{v-v_d}{v-v_s}\right)$

Where $f_d$ represents frequency perceived by the detector, $f_s$ represents wave’s frequency, $v$ represents speed of sound in air, $v_s$ represents velocity of source and $v_d$ represents velocity of detector.

Calculation:

As per problem,

The source train is moving away from the detector, so $v_s=-31.0\text{ m/s}$

The detector train is also moving away from the source train, so $v_d=-21.0\text{m/s}$

Now put the value of $f_s=305\text{ Hz}$ , $v=343\text{ m/s}$ , $v_d=-21.0\text{m/s}$ and $v_s=-31.0\text{ m/s}$ in the formula

  $\begin{array}{l}{f}_d=f_s\left(\frac{v-v_d}{v-v_s}\right)\\ f_d=305\text{ Hz}\left(\frac{343\text{ m/s}-(-21.0\text{ m/s})}{343\text{ m/s}-(-31.0\text{ m/s})}\right)\\ f_d=305\text{ Hz}\left(\frac{364\text{ m/s}}{374\text{ m/s}}\right)\\ f_d=297\text{ Hz}\end{array}$

Conclusion:

Thus, the frequency perceived by the detector train is $297\text{ Hz}$.