SATHEE: Simple Harmonic Motion 3 Question 5

Simple Harmonic Motion 3 Question 5

5. A simple pendulum has time period $T_{1}$ . The point of suspension is now moved upward according to the relation $y = k t^{2}, (k = 1 m / s^{2})$ , where $y$ is the vertical displacement.

The time period now becomes $T_{2}$ . The ratio of $\frac{T_{1}^{2}}{T_{2}^{2}}$ is

$(Take, g = 10 m / s^{2})$

(2005, 2M)

(a) $6 / 5$

(b) $5 / 6$

(d) $4 / 5$

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Answer:

Correct Answer: 5. (a)

Solution:

$y = k t^{2} \Rightarrow \frac{d^{2} y}{d t^{2}} = 2 k$

$\begin{aligned} or & a_{y} & = 2 m / s^{2} \\ T_{1} & = 2 π \sqrt{\frac{l}{g}} \\ and & T_{2} & = 2 π \sqrt{\frac{l}{g + a_{y}}} \\ ∴ & \frac{T_{1}^{2}}{T_{2}^{2}} & = \frac{g + a_{y}}{g} = \frac{10 + 2}{10} = \frac{6}{5} \end{aligned}$

$∴$ Correct answer is (a).

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Simple Harmonic Motion 3 Question 5

5. A simple pendulum has time period $T_{1}$ . The point of suspension is now moved upward according to the relation $y = k t^{2}, (k = 1 m / s^{2})$ , where $y$ is the vertical displacement.

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

Simple Harmonic Motion 3 Question 5

5. A simple pendulum has time period T1. The point of suspension is now moved upward according to the relation y=kt2,(k=1m/s2), where y is the vertical displacement.

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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5. A simple pendulum has time period $T_{1}$ . The point of suspension is now moved upward according to the relation $y = k t^{2}, (k = 1 m / s^{2})$ , where $y$ is the vertical displacement.