JEE Advanced 2018 · Paper 1 · Physics Q8

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PhysicsNumerical+3Question 8 of 54

Two men are walking along a horizontal straight line in the same direction. The man in front walks at a speed $1.0\ \mathrm{m\,s^{-1}}$ and the man behind walks at a speed $2.0\ \mathrm{m\,s^{-1}}$ . A third man is 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 $330\ \mathrm{m\,s^{-1}}$ . 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 ______.

Your answer

Answer and worked solution

Official answer: 5

Geometry. The stationary man is equidistant from the two, so he is above the midpoint. Each walker is 5 m away horizontally and $\sqrt{5^2+12^2}=13$ m away in all. So $\cos\theta=\dfrac5{13}$ .

Line-of-sight speeds. The rear man (2 m/s) approaches: $u_1=2\cdot\tfrac5{13}=\tfrac{10}{13}$ m/s. The front man (1 m/s) recedes: $u_2=\tfrac5{13}$ m/s.

Frequencies.
$f_1=f\frac{v}{v-u_1},\qquad f_2=f\frac v{v+u_2}.$

Beats. Since $u\ll v$ ,
$f_1-f_2\approx f\,\frac{u_1+u_2}v=1430\times\frac{15/13}{330}=1430\times\frac{15}{4290}=5\text{ Hz}.$

Answer: 5

Topics:Doppler effect Beats

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JEE Advanced 2018 · Paper 1, Physics question 8