Question

(a) What frequency is received by a person watching an oncoming ambulance moving at\(110\;{\rm{km/h}}\)and emitting a steady\(800\;{\rm{Hz}}\)sound from its siren? The speed of sound on this day is\(345\;{\rm{m/s}}\). (b) What frequency does she receive after the ambulance has passed?

Short answer

(a) The apparent frequency is \(877.58\;{\rm{Hz}}\).

(b) The apparent frequency is \(735.01\;{\rm{Hz}}\).

Step-by-step solution

The Concept of Apparent Frequency

The apparent frequency to a listener changes depending the sound source is moving away or toward the listener.

Given Data

The frequency is \(f = 800\;{\rm{Hz}}\).

The speed of the ambulance is \({v_s} = 100\;{\rm{km/h}} = 30.5\;{\rm{m/s}}\).

The speed of sound is \(v = 345\;{\rm{m/s}}\).

The speed of the listener is zero.

 Calculation of the frequency of the oncoming ambulance

(a)

The Doppler Effect tells the apparent frequency is,

\(f' = f\left( {\frac{{v - {v_o}}}{{v - {v_s}}}} \right)\)

Now, the apparent frequency is,

\(\begin{array}{c}f' = 800\left( {\frac{{345 - 0}}{{345 - 30.5}}} \right)\\ = 800 \times \frac{{690}}{{629}}\\ = 877.58\;{\rm{Hz}}\end{array}\)

Hence, the apparent frequency is \(877.58\;{\rm{Hz}}\) .

Calculation of the frequency of the passing away ambulance

(b)

Now, the apparent frequency of the passing away ambulance is,

\(\begin{array}{c}f'' = 800\left( {\frac{{345 - 0}}{{345 - \left( { - 30.5} \right)}}} \right)\\ = 800 \times \frac{{690}}{{751}}\\ = 735.01\;{\rm{Hz}}\end{array}\)

Hence, the apparent frequency is \(735.01\;{\rm{Hz}}\) .