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(a) At an air show a jet flies directly toward the stands at a speed of\({\rm{1200}}\;{\rm{km/h}}\), emitting a frequency of\({\rm{3500}}\;{\rm{Hz}}\), on a day when the speed of sound is\({\rm{342}}\;{\rm{m/s}}\). What frequency is received by the observers? (b) What frequency do they receive as the plane flies directly away from them?

Short Answer

Expert verified

(a) The apparent frequency is\(137586.2\;{\rm{Hz}}\).

(b) The apparent frequency is\(1772.5\;{\rm{Hz}}\).

Step by step solution

01

The relative motion and the frequency

The apparent frequency of a moving source increases when the source moves towards the listener. This change in apparent frequency is the Doppler effect.

02

Given Data

The frequency is\(f = 3500\;{\rm{Hz}}\).

The speed of the ambulance is\({v_s} = 1200\;{\rm{km/h}} = 333.3\;{\rm{m/s}}\).

The speed of sound is\(v = 342\;{\rm{m/s}}\).

The speed of the listener is zero.

03

Calculation of the frequency of the jet moving towards the observer

(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,

\(\begin{array}{c}f' = 3500\left( {\frac{{342 - 0}}{{342 - 333.3}}} \right)\\ = 3500 \times \frac{{1140}}{{29}}\\ = 137586.2\;{\rm{Hz}}\end{array}\)

Hence, the apparent frequency is\(137586.2\;{\rm{Hz}}\)

04

Calculation of the frequency of the jet moving away the observer

(b)

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

\(\begin{array}{c}f'' = 3500\left( {\frac{{342 - 0}}{{342 - \left( { - 333.3} \right)}}} \right)\\ = 3500 \times \frac{{1140}}{{2251}}\\ = 1772.5\;{\rm{Hz}}\end{array}\)

Hence, the apparent frequency is \(1772.5\;{\rm{Hz}}\)

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