SATHEE: Wave Motion 2 Question 46

Wave Motion 2 Question 46

46. The vibrations of a string of length $60 c m$ fixed at both ends are represented by the equation

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Answer:

Correct Answer: 46. (a) $2 \sqrt{3} c m$ (b) $x = 0, 15 c m, 30 c m \dots$ etc

(d) $y_{1} = 2 \sin (\frac{π x}{15} - 96 π t)$ and $y_{2} = 2 \sin (\frac{π x}{15} + 96 π t)$

Solution:

(a) $y = 4 \sin \frac{π x}{15} \cos (96 π t) = A_{x} \cos (96 π t)$

Here, $A_{x} = 4 \sin \frac{π x}{15}$

at $x = 5 c m, A_{x} = 4 \sin \frac{5 π}{15} = 4 \sin \frac{π}{3} = 2 \sqrt{3} c m$

This is the amplitude or maximum displacement at

$x = 5 c m .$

(b) Nodes are located where $A_{x} = 0$

$or \frac{π x}{15} = 0, π, 2 π \dots \dots or x = 0, 15 c m, 30 c m e t c .$

$v_{p} = {| \frac{\partial y}{\partial t} |}_{x = constant} = - 384 π \sin (\frac{π x}{15}) \sin (96 π t)$

At $x = 7.5 c m$ and $t = 0.25 s$

$v_{p} = - 384 π \sin (\frac{π}{2}) \sin (24 π) = 0$

(d) Amplitude of components waves is $A = \frac{4}{2} = 2 c m$

$ω = 96 π and k = \frac{π}{15}$

$∴$ Component waves are,

and

$\begin{aligned} y_{1} = 2 \sin (\frac{π}{15} x - 96 π t) \\ y_{2} = 2 \sin (\frac{π}{15} x + 96 π t) \end{aligned}$

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Wave Motion 2 Question 46

46. The vibrations of a string of length $60 c m$ fixed at both ends are represented by the equation

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

Wave Motion 2 Question 46

46. The vibrations of a string of length 60cm fixed at both ends are represented by the equation

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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46. The vibrations of a string of length $60 c m$ fixed at both ends are represented by the equation