Question

A boy is walking away from a well at a speed of $1.0\,{\text{m/s}}$ in a direction at right angles to the wall. AS he walks, he blows a whistle steadily. An observer towards whom the boy is walking hears 4.0 beats per second. If the speed of sound is $340\,{\text{m/s}}$, what is the frequency of the whistle (in Hz) ?

Answer 1

Use the formula for the frequency of the sound heard by a stationary observer from a source approaching towards the observer. This formula gives the relation between the frequency heard by an observer, frequency of the source, velocity of the sound and velocity of the source. Also determine the frequency heard by the observer after reflection of sound from the wall.

Formula used:
The frequency of a sound heard by a stationary observer from a source approaching towards the observer is given by
$f' = f\left( {\dfrac{v}{{v - {v_S}}}} \right)$ …… (1)
Here, $f'$ is the frequency of the sound heard by the observer, $f$ is the frequency of the sound from the source, $v$ is the velocity of the sound and ${v_S}$ is the velocity of the source.

Complete step by step answer:
We have given that a boy blowing whistle is walking away from a wall at a speed of $1.0\,{\text{m/s}}$ towards an observer.
${v_S} = 1.0\,{\text{m/s}}$

It is given that the observer is stationary.

The speed of the sound is $340\,{\text{m/s}}$.
$v = 340\,{\text{m/s}}$

We can determine the frequency of the whistle heard by the stationary observer from the boy directly using equation (1).

Substitute $1\,{\text{m/s}}$ for ${v_S}$ and $340\,{\text{m/s}}$ for $v$ in equation (1).
$f' = f\left( {\dfrac{{340\,{\text{m/s}}}}{{\left( {340\,{\text{m/s}}} \right) - \left( {1\,{\text{m/s}}} \right)}}} \right)$

The stationary observer also hears the sound of the whistle after the reflection from the wall.

Rewrite equation (2) for the frequency $f''$ of whistle heard by the boy after reflection from the wall.
$f'' = f\left( {\dfrac{v}{{v + {v_S}}}} \right)$

Substitute $1\,{\text{m/s}}$ for ${v_S}$ and $340\,{\text{m/s}}$ for $v$ in the above equation.
$f'' = f\left( {\dfrac{{340\,{\text{m/s}}}}{{\left( {340\,{\text{m/s}}} \right) + \left( {1\,{\text{m/s}}} \right)}}} \right)$

The beat frequency $\Delta f$ heard by the observer is $4\,{\text{beats}} \cdot {{\text{s}}^{ - 1}}$.
$\Delta f = 4\,{\text{beats}} \cdot {{\text{s}}^{ - 1}}$

The beat frequency is the difference between the two frequencies heard by the stationary observer.
$\Delta f = f' - f''$

Substitute $4\,{\text{beats}} \cdot {{\text{s}}^{ - 1}}$ for $\Delta f$, $f\left( {\dfrac{{340\,{\text{m/s}}}}{{\left( {340\,{\text{m/s}}} \right) - \left( {1\,{\text{m/s}}} \right)}}} \right)$ for $f'$ and $f\left( {\dfrac{{340\,{\text{m/s}}}}{{\left( {340\,{\text{m/s}}} \right) + \left( {1\,{\text{m/s}}} \right)}}} \right)$ for $f''$ in the above equation.
$4\,{\text{beats}} \cdot {{\text{s}}^{ - 1}} = f\left( {\dfrac{{340\,{\text{m/s}}}}{{\left( {340\,{\text{m/s}}} \right) - \left( {1\,{\text{m/s}}} \right)}}} \right) - f\left( {\dfrac{{340\,{\text{m/s}}}}{{\left( {340\,{\text{m/s}}} \right) + \left( {1\,{\text{m/s}}} \right)}}} \right)$
$\Rightarrow 4\, = f\left( {\dfrac{{340}}{{339}}} \right) - f\left( {\dfrac{{340}}{{341}}} \right)$
$\Rightarrow 4\, = 340f\left( {\dfrac{1}{{339}} - \dfrac{1}{{341}}} \right)$
$\Rightarrow 4\, = 340f\left( {\dfrac{{341 - 339}}{{339 \times 341}}} \right)$
$\Rightarrow 4\, = 340f\left( {\dfrac{2}{{339 \times 341}}} \right)$
$\Rightarrow f = \dfrac{{4\, \times 339 \times 341}}{{2 \times 340}}$
$\Rightarrow f = 679.9\,{\text{Hz}}$
$\therefore f \approx 680\,{\text{Hz}}$

Hence, the frequency of the whistle is $680\,{\text{Hz}}$.

Note: In the formula for the frequency heard by the boy after reflection of the sound from the wall, the sign between two velocities in the denominator is changed from minus to plus. This is because the direction of the sound and the source becomes opposite in this case as the boy is moving away from the wall and the sound approaches the wall.