Question

A boy is walking away from a wall at a speed of $1.0m{s^{ - 1}}$ 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 $340m{s^{ - 1}}$, what is the frequency of the whistle (in $Hz$)?

Answer 1

Here, you have a situation in which the boy is walking away from the wall and towards the observer and is whistling steadily. You are asked to find the frequency of the whistle. In order to find the frequency of whistles, the first thing you need to figure out is why the observer is able to hear beats. There is only one source of sound and that is the whistle, you need to find the other source. After understanding the situation properly, use the Doppler Effect to find the frequency of the whistle. Also, use the basic definition of beats to obtain equations.

Complete step by step answer:
As you know that the observer is hearing beats, there are supposed to be two frequencies, let those frequencies be ${f_1}$ and ${f_2}$. Now, will look at the question carefully and then determine ${f_1}$ and ${f_2}$. It is given to you that the boy is moving towards the observer whistling, so one of the frequencies is the one that is reaching the observer directly, but you are also given that the boy is moving away from the wall, which means that the second frequency is the frequency which corresponds to the event when the sound wave travels to the wall, reflects and then reaches the observer.

This is how two different frequencies are created. As you know that beats are defined as differences between frequencies, we have the beats heard by observers as ${f_1} - {f_2}$. Now, it is intuitive that ${f_1}$ will be the frequency which is directly reaching the observer as it will be larger and ${f_2}$ will be the reflected one.Now, let us use the Doppler Effect. The frequency directly reaching the observer will be given as,
${f_1} = \left( {\dfrac{{u \pm {v_o}}}{{u \pm {v_s}}}} \right){f_0}$, where $u$ is the speed of sound, ${v_0}$ is the speed of observer,${v_s}$ is the speed of source and ${f_0}$ is the actual frequency. In our case, ${v_o} = 0$, $u = 340m{s^{ - 1}}$ and ${v_s} = 1m{s^{ - 1}}$.

${f_1} = \left( {\dfrac{{340}}{{340 - 1}}} \right){f_0} = \dfrac{{340{f_0}}}{{339}}$
We took the negative because of the sign convention, as it is moving towards the observer, so the frequency received would be higher than the actual frequency.Now, we will calculate the frequency perceived by the wall because due to reflection, that is the frequency that will be perceived by the observer. In this case, we have, ${v_o} = 0$ and ${v_s} = 1m{s^{ - 1}}$. Here, we will use positive sign due to sign convention because the source is receding from the observer. It will be given as,
${f_2} = \left( {\dfrac{{340}}{{340 + 1}}} \right){f_0} = \dfrac{{340{f_0}}}{{341}}$
Let us substitute these values in the equation of beats, we get,
${f_1} - {f_2} = 4 \\\ \Rightarrow\dfrac{{340{f_0}}}{{339}} - \dfrac{{340{f_0}}}{{341}} = 4 \\\ \therefore{f_0} = 679.99Hz \\\$
Therefore, the frequency of the whistle is $680Hz$.

Note: Here, you should pay focus on the sources of the frequencies by which the beats are produced. Also, keep in mind the sign convention, when the observer is moving towards the source, positive sign of observer is taken, else negative and when the source is moving towards the observer, negative sign is taken, else positive. Remember the Doppler Effect and also the definition of beats.