What is the minimum speed at which a source must travel toward you for you to be able to hear that its frequency is Doppler shifted? That is, what speed produces a shift of\({\rm{0}}{\rm{.300\% }}\)on a day when the speed of sound is\({\rm{331}}\;{\rm{m/s}}\)?

The speed of sound is\(v = 331\;{\rm{m/s}}\).

The speed of the source is\({v_s} = 10\;{\rm{m/s}}\).

The apparent frequency of sound changes with the relative velocity of the source and the listener.

According to the Doppler Effect the apparent frequency is,

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

Now, the source is coming towards the observer, so all the quantities are positive.

So, for the given values,

\(\begin{array}{c}\frac{{f' - f}}{f} = \frac{{{\rm{0}}{\rm{.3}}}}{{{\rm{100}}}}\\\frac{{f'}}{f} = {\rm{1}}{\rm{.003}}\end{array}\)

Substituting the values,

\(\begin{array}{c}\frac{{f'}}{f} = \left( {\frac{{{\rm{331 - 0}}}}{{{\rm{331 - }}{{\rm{v}}_{\rm{s}}}}}} \right)\\{\rm{1}}{\rm{.003 = }}\frac{{{\rm{331}}}}{{{\rm{331}} - {v_s}}}\\{\rm{331}} - {v_s}{\rm{ = }}\frac{{{\rm{331}}}}{{{\rm{1}}{\rm{.003}}}}\\{v_s} = {\rm{1}}\;{\rm{m/s}}\end{array}\)

Now, plugging the values for the faster eagle,

\(\begin{array}{c}f' = 3200\left( {\frac{{330 - 20}}{{330 - 15}}} \right)\\ = 3200 \times \frac{{310}}{{315}}\\ = 3739.68\;{\rm{Hz}}\end{array}\)

Hence, the minimum speed of the source is \({\rm{1}}\;{\rm{m/s}}\).