Question

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\({\rm{331}}\;{\rm{m/s}}\)?

Short answer

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

Step-by-step solution

Given Data

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

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

Doppler Effect

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

Calculation of the Apparent Frequency

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 \({\rm{1}}\;{\rm{m/s}}\).