Question

One foggy morning, when the temperature of t he air was 19 degrees calcius, Kenny is driving his speed boat toward the Brant Point Lighthouse at a speed of 15 m/s as the fog horn blows with a frequency of 180 Hz. What frequency does Kenny hear as he moves?

Answer 1

Concepts and reason

The concept used in this question is Doppler Effect. The frequency heard by Kenny can be calculated using the formula of Doppler Shift.

Fundamentals

Doppler Effect:

The Doppler Effect is the change in frequency or wavelength due to a moving wave relative to the observer who is also moving. The shift in frequency is called Doppler Shift.

The formula for Doppler Effect is:

${f_{\rm{o}}} = f\frac{{v \pm {v_{\rm{o}}}}}{{v \pm {v_{\rm{s}}}}}$

Here, $f$ is the frequency of original frequency, ${f_{\rm{o}}}$ is the frequency of observer, $v$ is the speed of sound, ${v_{\rm{o}}}$ is the speed of observer and ${v_{\rm{s}}}$ is the speed of source.

If the source and observer are moving towards each other, then the observed frequency is:

${f_{\rm{o}}} = f\frac{{v + {v_{\rm{o}}}}}{{v - {v_{\rm{s}}}}}$ …… (1)

If the source and observer are moving away from each other, then the observed frequency is:

${f_{\rm{o}}} = f\frac{{v - {v_{\rm{o}}}}}{{v + {v_{\rm{s}}}}}$ …… (2)

Here, the source is lighthouse and the observer is Kenny. Kenny moves into the fog and the lighthouse is stationary.

So, the formula to be used here for calculating the frequency observed by Kenny must be:

${f_{\rm{o}}} = f\frac{{v + {v_{\rm{o}}}}}{{v - {v_{\rm{s}}}}}$

Here, $f$ is the frequency of fog, ${f_{\rm{o}}}$ is the frequency observed by Kenny, $v$ is the speed of sound, ${v_{\rm{o}}}$ is the speed of Kenny and ${v_{\rm{s}}}$ is the speed of lighthouse.

The observed frequency is:

${f_{\rm{o}}} = f\frac{{v + {v_{\rm{o}}}}}{{v - {v_{\rm{s}}}}}$

Substitute $180{\rm{ Hz}}$ for $f$ , $340{\rm{ m/s}}$ $v$ , $15{\rm{ m/s}}$ for ${v_{\rm{o}}}$ and $0{\rm{ m/s}}$ for ${v_{\rm{s}}}$ .

$\begin{array}{c}\\{f_{\rm{o}}} = \left( {180{\rm{ Hz}}} \right)\frac{{\left( {340{\rm{ m/s}}} \right) + \left( {15{\rm{ m/s}}} \right)}}{{\left( {340{\rm{ m/s}}} \right) - \left( {0{\rm{ m/s}}} \right)}}\\\\ = \left( {180{\rm{ Hz}}} \right)\frac{{355{\rm{ m/s}}}}{{340{\rm{ m/s}}}}\\\\ = \left( {180{\rm{ Hz}}} \right)\left( {1.044} \right)\\\\ = {\bf{188 Hz}}\\\end{array}$

Ans:

The frequency heard by Kenny is ${\bf{188 Hz}}$ .