SATHEE: Simple Harmonic Motion 5 Question 5

Simple Harmonic Motion 5 Question 5

10. A point mass is subjected to two simultaneous sinusoidal displacements in $x$ -direction, $x_{1} (t) = A \sin ω t$ and

$x_{2} (t) = A \sin ω t + \frac{2 π}{3}$ . Adding a third sinusoidal displacement $x_{3} (t) = B \sin (ω t + φ)$ brings the mass to a complete rest. The values of $B$ and $φ$ are

(2011)

(a) $\sqrt{2} A, \frac{3 π}{4}$

(b) $A, \frac{4 π}{3}$

(d) $A, \frac{π}{3}$

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Answer:

Correct Answer: 10. (b)

Solution:

Resultant amplitude of $x_{1}$ and $x_{2}$ is $A$ at angle $\frac{π}{3}$ from $A_{1}$ . To make resultant of $x_{1}, x_{2}$ and $x_{3}$ to be zero. $A_{3}$ should be equal to $A$ at angle $φ = \frac{4 π}{3}$ as shown in figure.

$∴$ Correct answer is (b).

Alternate Solution.

It we substitute, $x_{1} + x_{2} + x_{3} = 0$

or $A \sin ω t + A \sin ω t + \frac{2 π}{3} + B \sin (ω t + φ) = 0$

Then by applying simple mathematics we can prove that

$B = A and φ = \frac{4 π}{3}$

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अगला

Simple Harmonic Motion 5 Question 5

10. A point mass is subjected to two simultaneous sinusoidal displacements in $x$ -direction, $x_{1} (t) = A \sin ω t$ and

Answer:

Alternate Solution.

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Simple Harmonic Motion 5 Question 5

10. A point mass is subjected to two simultaneous sinusoidal displacements in x-direction, x1(t)=Asin⁡ωt and

Answer:

Alternate Solution.

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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10. A point mass is subjected to two simultaneous sinusoidal displacements in $x$ -direction, $x_{1} (t) = A \sin ω t$ and