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QJEEAdv18Phf86a: JEE Advanced 2018 Physics - Class 11 JEEAdv18Phf86a Solution: Two men are walking along a horizontal straight line in the…

Two men are walking along a horizontal straight line in the same direction. The man in front walks at a speed 1.0 ms−11.0\,m{s^{ - 1}} and the man behind walks at a speed 2.0 ms−1.2.0\,m{s^{ - 1}}. A third man in standing at a height 1212 mm 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 14301430 HzHz . The speed of sound in air is 330 ms−1.330\,m{s^{ - 1}}. At the instant, when the moving men are 1010 mm apart, the stationary man is equidistant from them. The frequency of beats in Hz,Hz, heard by the stationary man at this instant, is _____________.

Step-by-Step Verified Solution

From given data, we can draw the figure below.

JEE Advanced 2018 Paper 1 Offline Physics - Waves Question 32 English Explanation

Now, frequency of man behind is

f1=f(vv−vscos⁡θ){f_1} = f\left( {{v \over {v - {v_s}\cos \theta }}} \right)

where, x is speed of sound in air, vs is speed of man and f is frequency of whistle. Therefore,

f1=1430(330330−2×513){f_1} = 1430\left( {{{330} \over {330 - 2 \times {5 \over {13}}}}} \right)

Frequency of man in front is

f2=f(vv+vscos⁡θ)=1430(330330+1×513){f_2} = f\left( {{v \over {v + {v_s}\cos \theta }}} \right) = 1430\left( {{{330} \over {330 + 1 \times {5 \over {13}}}}} \right)

Now, frequency of beat is given as

Δf=f1−f2\Delta f = {f_1} - {f_2}

=1430(330330−1013)−1430(330330−513)= 1430\left( {{{330} \over {330 - {{10} \over {13}}}}} \right) - 1430\left( {{{330} \over {330 - {5 \over {13}}}}} \right)

=1430×330([1330−1013]−[1330−513])= 1430 \times 330\left( {\left[ {{1 \over {330 - {{10} \over {13}}}}} \right] - \left[ {{1 \over {330 - {5 \over {13}}}}} \right]} \right)

=1430×330[134280−134295]= 1430 \times 330\left[ {{{13} \over {4280}} - {{13} \over {4295}}} \right]

=1430×330×13×154280×4295= {{1430 \times 330 \times 13 \times 15} \over {4280 \times 4295}} = 5.0058 Hz ≈\approx 5.00 Hz

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