Question 41: Waves

Question

Two men are walking along a horizontal straight line in the same direction. The man in front walks at a speed $$1.0\,m{s^{ - 1}}$$ and the man behind walks at a speed $$2.0\,m{s^{ - 1}}.$$ A third man in standing at a height $$12$$ $$m$$ above the same horizontal line such that all three men are in a vertical  plane. The two walking men are blowing identical whistles which emit a sound of frequency $$1430$$ $$Hz$$.  The speed of sound in air is $$330\,m{s^{ - 1}}.$$ At the instant, when the moving men are $$10$$ $$m$$ apart, the stationary man is equidistant from them. The frequency of beats in $$Hz,$$ heard by the stationary man at this instant, is _____________.

Accepted Answer

From given data, we can draw the figure below.
Now, frequency of man behind is
$${f_1} = f\left( {{v \over {v - {v_s}\cos 	heta }}} ight)$$
where, x is speed of sound in air, v
s
is speed of man and f is frequency of whistle. Therefore,
$${f_1} = 1430\left( {{{330} \over {330 - 2 	imes {5 \over {13}}}}} ight)$$
Frequency of man in front is
$${f_2} = f\left( {{v \over {v + {v_s}\cos 	heta }}} ight) = 1430\left( {{{330} \over {330 + 1 	imes {5 \over {13}}}}} ight)$$
Now, frequency of beat is given as
$$\Delta f = {f_1} - {f_2}$$
$$ = 1430\left( {{{330} \over {330 - {{10} \over {13}}}}} ight) - 1430\left( {{{330} \over {330 - {5 \over {13}}}}} ight)$$
$$ = 1430 	imes 330\left( {\left[ {{1 \over {330 - {{10} \over {13}}}}} ight] - \left[ {{1 \over {330 - {5 \over {13}}}}} ight]} ight)$$
$$ = 1430 	imes 330\left[ {{{13} \over {4280}} - {{13} \over {4295}}} ight]$$
$$ = {{1430 	imes 330 	imes 13 	imes 15} \over {4280 	imes 4295}}$$ = 5.0058 Hz $$\approx$$ 5.00 Hz

Waves

Choose the correct answer: