SATHEE: Wave Motion 4 Question 19

Wave Motion 4 Question 19

18. Two men are walking along a horizontal straight line in the same direction. The main 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 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 $1430 H z$ . The speed of sound in air $330 m s^{- 1}$ . At the instant, when the moving men are $10 m$ apart, the stationary man is equidistant from them. The frequency of beats in $H z$ , heard by the stationary man at this instant, is

(2018 Adv.)

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Answer:

Correct Answer: 18. (b)

Solution:

$\begin{aligned} f_{A} & = 1430 [\frac{330}{330 - 2 \cos θ}] \\ = 1430 [\frac{1}{1 - \frac{2 \cos θ}{330}}] \approx 1430 [1 + \frac{2 \cos θ}{330}] \end{aligned}$

[from binomial expansion]

$f_{B} = 1430 [\frac{330}{330 + 1 \cos θ}] \approx 1430 [1 - \frac{\cos θ}{330}]$

$\begin{aligned} Beat frequency & = f_{A} - f_{B} = 1430 [\frac{3 \cos θ}{330}] = 13 \cos θ \\ = 13 (\frac{5}{13}) = 5.00 H z \end{aligned}$

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Wave Motion 4 Question 19

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Answer:

इंजीनियरिंग की तैयारी

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Wave Motion 4 Question 19

Fill in the Blank

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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