A spectator at a parade receives an\(888\;{\rm{Hz}}\)tone from an oncoming trumpeter who is playing an\(880\;{\rm{Hz}}\)note. At what speed is the musician approaching if the speed of sound is\({\rm{338}}\;{\rm{m/s}}\)?

The apparent frequency of a sound source increases when the source moves towards the listener and vise-versa.

The apparent frequency is \(f' = 888\;{\rm{Hz}}\).

The actual frequency is \(f = 880\;{\rm{Hz}}\).

The speed of sound is \(v = 338\;{\rm{m/s}}\).

The Doppler Effect tells the apparent frequency is,

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

Now, the apparent frequency is,

\(\begin{array}{c}888 = 880\left( {\frac{{338 - 0}}{{338 - {v_s}}}} \right)\\\frac{{888}}{{880}} = \frac{{338}}{{338 - {v_s}}}\\338 - {v_s} = 334.95\\{v_s} = 3.05\;{\rm{m/s}}\end{array}\)

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, thespeed of the trumpeter is \(3.05\;{\rm{m/s}}\).