Question

Two men are walking along a horizontal straight line in the same direction. The man in the front walks at a speed $2\text{m/s}$ and the man behind walks at a speed $4\text{m/s}$. A third man is standing at a height $24\text{m}$ above the same horizontal line such that all three men are in the same vertical plane. The two walking men are blowing identical whistles which emit sounds of frequency $2010\text{Hz}$. The speed of sound in air is $335\text{m/s}$. At the instant when the moving men are $14\text{m}$ apart, the stationary man is equidistant from them. The number of beats per second heard by the stationary man is

(give your answer as the nearest integer value)

Answer 1

When the stationary man is equidistant from the two moving men (diagram shown below):


Given that,

Actual frequency heard by the observer at rest when source is also at rest is given by
$f_0=2010\text{Hz}$

Distance, $AB=14\text{m}$
So, $AP=PB=7\text{m}$

Height $\left(OP\right)=24\text{m}$

In $\Delta OAP$

$OA=\sqrt{{\left(AP\right)}^2+{\left(OP\right)}^2}=\sqrt{7^2+{24}^2}=25\text{m}$

and, $\cos \theta =\frac{7}{25}$

Now for source A, frequency received by stationary observer

$f_A=f_0\left[\frac{V}{V-V_{AO}\cos \theta }\right]$

[where $V$ is speed of sound and $V_{AO}$ is speed of A w.r.t observer O ]

Substituting the given data,

$f_A=2010\left[\frac{335}{335-4\times \frac{7}{25}}\right]\approx 2017\text{Hz}$

similarly for source B,

$f_B=f_0\left[\frac{V}{V+V_{BO}\cos \theta }\right]$

[where $V$ is speed of sound and $V_{BO}$ is speed of B w.r.t observer O ]

Substituting the given data,

$f_B=2010\left[\frac{335}{335+2\times \frac{7}{25}}\right]\approx 2007\text{Hz}$

So, beat frequency heard by stationary observer

$f=f_A-f_B=2017-2007=10\text{Hz}$
i.e 10 beats per second.