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PhysicsSHMCalculation of time periodEasy2 minQB

PhysicsEasysingle choice

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 $(g = 10 m s^{- 2})$

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Answer:

Solution:

Given, Length of simple pendulum, $l = \sqrt{5} m$

Acceleration of car $(a) = 5 m s^{- 2}$

Now $g_{e f f e c t i v e} = \sqrt{a^{2} + g^{2}}$

$= \sqrt{{(5)}^{2} + {(10)}^{2}}$

$= \sqrt{125} = 5 \sqrt{5}$

$T = 2 π \sqrt{\frac{l}{g_{e f f e c t i v e}}}$

$T = 2 π \sqrt{\frac{\sqrt{5}}{5 \sqrt{5}}}$

$T = \frac{2 π}{\sqrt{5}} s$

Stream:JEESubject:PhysicsTopic:SHMSubtopic:Calculation of time period

⏱ 2mℹ️ Source: QB

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