Q60P (II) In one of the original Dopp... [FREE SOLUTION]

Chapter 12: Q60P (page 328)

(II) In one of the original Doppler experiments, a tuba was played at a frequency of 75 Hz on a moving flat train car, and a second identical tuba played the same tone while at rest in the railway station. What beat frequency was heard in the station if the train car approached the station at a speed of 14.0 m/s?

Short Answer

Expert verified

The beat frequency is \(3.12\;{\rm{Hz}}\).

Step by step solution

Understanding the frequency of the observer

In this condition, the stationary tuba (observer) will obtain the frequency from the moving tuba when the train car approaches the station.

Given Data

The frequency of tuba is\(f = 75\;{\rm{Hz}}\).

The speed of the train is \(v = 14\;{\rm{m/s}}\).

Evaluating the beat frequency

The relation of frequency is given by,

\({f_0} = f\left( {\frac{1}{{1 - \frac{v}{{{v_{\rm{s}}}}}}}} \right)\)

Here, \({v_{\rm{s}}}\) is the speed of the sound.

On plugging the values in the above relation.

\(\begin{aligned}{l}{f_{\rm{0}}} &= \left( {75\;{\rm{Hz}}} \right)\left( {\frac{1}{{1 - \frac{{14\;{\rm{m/s}}}}{{343\;{\rm{m/s}}}}}}} \right)\\{f_{\rm{0}}} &= 78.12\;{\rm{Hz}}\end{aligned}\)

The relation to the calculate beat frequency is given by,

\(\begin{aligned}{c}{f_{\rm{b}}} &= \left| {{f_0} - f} \right|\\{f_{\rm{b}}} &= \left| {78.12\;{\rm{Hz}} - 75\;{\rm{Hz}}} \right|\\{f_{\rm{b}}} &= 3.12\;{\rm{Hz}}\end{aligned}\)

Thus, \({f_{\rm{b}}} = 3.12\;{\rm{Hz}}\) is the required beat frequency.