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)
B
X2/(a2X2)
C
a2X2ω2/X2X2ω2
D
(a2X2)/X2

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 answerPhysics

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