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Question

The angular velocity and the amplitude of simple pendulum is $$\omega$$ and $$a$$ respectively. At a displacement $$X$$ from mean position, if the kinetic energy is $$T$$ and potential energy is $$V$$. Then ratio $$T$$ to $$V$$


A
X2ω2/(a2X2ω2)
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B
X2/(a2X2)
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C
a2X2ω2/X2X2ω2
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D
(a2X2)/X2
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Solution

The correct option is A $$(a^{2} - X^{2})/X^2$$
We know,
$$T=\dfrac{1}{2}m\omega^2(a^2-X^2)$$

$$V=\dfrac{1}{2}m\omega^2X^2$$

$$T:V=\dfrac{a^2-X^2}{X^2}$$

Option $$\textbf D$$ is the correct answer

Physics

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