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Waves question

2018 · Shift 1 · Q41
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Waves question

2018 · Shift 1 · Q41

JEE AdvancedPhysicsWavesNumerical+3 / −1
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}}1.0ms−1 and the man behind walks at a speed 2.0 ms−1.2.0\,m{s^{ - 1}}.2.0ms−1. A third man in standing at a height 12m12m12m 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 1430Hz1430Hz1430Hz. The speed of sound in air is 330 ms−1.330\,m{s^{ - 1}}.330ms−1. At the instant, when the moving men are 10m10m10m apart, the stationary man is equidistant from them. The frequency of beats in Hz,Hz,Hz, heard by the stationary man at this instant, is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 5

  1. Set up the geometry

Since the stationary man is at height 12 m12\,\text{m}12m above the horizontal line and is equidistant from the two moving men when they are 10 m10\,\text{m}10m apart, he must lie vertically above the midpoint of the segment joining them.

So each moving man is horizontally 5 m5\,\text{m}5m from the point directly below the stationary man.

Hence the distance from each whistle to the stationary observer is

r=122+52=144+25=13 m.r=\sqrt{12^2+5^2}=\sqrt{144+25}=13\,\text{m}.r=122+52​=144+25​=13m.
  1. Find the radial velocity component for each man

For Doppler effect, only the component of source velocity along the line joining source and observer matters.

The cosine of the angle between the horizontal direction of motion and the line joining source to observer is

cos⁡θ=513.\cos\theta=\frac{5}{13}.cosθ=135​.
  • The man in front moves at 1 m s−11\,\text{m s}^{-1}1m s−1.
  • The man behind moves at 2 m s−12\,\text{m s}^{-1}2m s−1.

Since both men are walking in the same direction, and the observer is above the midpoint:

  • the front man is moving away from the observer along the line of sight,
  • the rear man is moving towards the observer along the line of sight.

Therefore radial components are:

vr1=1⋅513=513 m s−1v_{r1}=1\cdot \frac{5}{13}=\frac{5}{13}\,\text{m s}^{-1}vr1​=1⋅135​=135​m s−1

(away), and

vr2=2⋅513=1013 m s−1v_{r2}=2\cdot \frac{5}{13}=\frac{10}{13}\,\text{m s}^{-1}vr2​=2⋅135​=1310​m s−1

(towards).

  1. Observed frequencies from the two whistles

For a moving source and stationary observer:

  • if source approaches,
f′=fvv−vsf'=f\frac{v}{v-v_s}f′=fv−vs​v​
  • if source recedes,
f′=fvv+vsf'=f\frac{v}{v+v_s}f′=fv+vs​v​

where v=330 m s−1v=330\,\text{m s}^{-1}v=330m s−1.

So,

  • For the front man (receding):
f1=1430⋅330330+513f_1=1430\cdot \frac{330}{330+\frac{5}{13}}f1​=1430⋅330+135​330​
  • For the rear man (approaching):
f2=1430⋅330330−1013f_2=1430\cdot \frac{330}{330-\frac{10}{13}}f2​=1430⋅330−1310​330​
  1. Compute the beat frequency

Beat frequency is

∣f2−f1∣.|f_2-f_1|.∣f2​−f1​∣.

Now,

330+513=4290+513=429513,330+\frac{5}{13}=\frac{4290+5}{13}=\frac{4295}{13},330+135​=134290+5​=134295​,

so

f1=1430⋅3304295/13=1430⋅42904295.f_1=1430\cdot \frac{330}{4295/13} =1430\cdot \frac{4290}{4295}.f1​=1430⋅4295/13330​=1430⋅42954290​.

Also,

330−1013=4290−1013=428013,330-\frac{10}{13}=\frac{4290-10}{13}=\frac{4280}{13},330−1310​=134290−10​=134280​,

so

f2=1430⋅3304280/13=1430⋅42904280.f_2=1430\cdot \frac{330}{4280/13} =1430\cdot \frac{4290}{4280}.f2​=1430⋅4280/13330​=1430⋅42804290​.

Thus

∣f2−f1∣=1430⋅4290(14280−14295).|f_2-f_1|=1430\cdot 4290\left(\frac{1}{4280}-\frac{1}{4295}\right).∣f2​−f1​∣=1430⋅4290(42801​−42951​).

Now,

14280−14295=4295−42804280⋅4295=154280⋅4295.\frac{1}{4280}-\frac{1}{4295}=\frac{4295-4280}{4280\cdot 4295} =\frac{15}{4280\cdot 4295}.42801​−42951​=4280⋅42954295−4280​=4280⋅429515​.

Hence

∣f2−f1∣=1430⋅4290⋅154280⋅4295.|f_2-f_1|=1430\cdot 4290\cdot \frac{15}{4280\cdot 4295}.∣f2​−f1​∣=1430⋅4290⋅4280⋅429515​.

This gives approximately

∣f2−f1∣≈5.0 Hz.|f_2-f_1|\approx 5.0\,\text{Hz}.∣f2​−f1​∣≈5.0Hz.
  1. Final answer

The frequency of beats heard is

5\boxed{5}5​
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