The time-period of a simple pendulum of length $\sqrt{5}$ m suspended in a car moving with uniform acceleration of $5 m s^{- 2}$ in a horizontal straight road is $(g = 10 m s^{- 2})$ :

\frac{2 π}{\sqrt{5}} s

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\frac{π}{\sqrt{5}} s

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5 π s

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4 π s

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3 π s

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Solution

The correct option is A $\frac{2 π}{\sqrt{5}} s$
Given, length of simple pendulum, $l = \sqrt{5}$ m
Acceleration of car (a) $= 5 m s^{- 2}$
Now $g_{e f f e c t i v e} = \sqrt{a^{2} + g^{2}}$
$g_{e f f e c t i v e} = \sqrt{(5)^{2} + (10)^{2}}$
$= \sqrt{125} = 5 \sqrt{5}$
$T = 2 π \sqrt{\frac{l}{g_{e f f e c t i v e}}}$
$T = 2 π \sqrt{\frac{\sqrt{5}}{5 \sqrt{5}}}$
$T = \frac{2 π}{\sqrt{5}} s$ .

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