A small sphere oscillates simple harmonically | vedclass

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(A) The time period $T$ of a simple pendulum is given by the formula $T = 2 \pi \sqrt{\frac{L}{g}}$.In the case of a sphere oscillating in a watch gla

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$0.8$

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(A) The time period $T$ of a simple pendulum is given by the formula $T = 2 \pi \sqrt{\frac{L}{g}}$.In the case of a sphere oscillating in a watch glass,the effective length $L$ of the pendulum is equal to the radius of curvature $R$ of the watch glass.Given,$R = L = 1.6 \ m$ and $g = 10 \ m/s^2$.Substituting these values into the formula:$T = 2 \pi \sqrt{\frac{1.6}{10}}$$T = 2 \pi \sqrt{0.16}$$T = 2 \pi 	imes 0.4$$T = 0.8 \pi \ s$.

$A$ small sphere oscillates simple harmonically in a watch glass whose radius of curvature is $1.6 \ m$. The period of oscillation of the sphere in seconds is (acceleration due to gravity,$g = 10 \ m/s^2$). (in $\pi$)

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