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(A) With respect to the cart,the equilibrium position of the pendulum is shifted due to the pseudo force acting on the bob. The effective acceleration

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(A) With respect to the cart,the equilibrium position of the pendulum is shifted due to the pseudo force acting on the bob. The effective acceleration $g_{	ext{eff}}$ acting on the pendulum is the vector sum of the acceleration due to gravity $g$ and the pseudo acceleration $a = \sqrt{3}g$ in the opposite direction of the cart's motion.$g_{	ext{eff}} = \sqrt{g^2 + a^2} = \sqrt{g^2 + (\sqrt{3}g)^2} = \sqrt{g^2 + 3g^2} = \sqrt{4g^2} = 2g$.Given $g = \pi^2 \ m/s^2$,we have $g_{	ext{eff}} = 2\pi^2 \ m/s^2$.The time period $T$ of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g_{	ext{eff}}}}$.Substituting $L = 50 \ cm = 0.5 \ m$ and $g_{	ext{eff}} = 2\pi^2 \ m/s^2$:$T = 2\pi \sqrt{\frac{0.5}{2\pi^2}} = 2\pi \sqrt{\frac{0.5}{2} \cdot \frac{1}{\pi^2}} = 2\pi \cdot \frac{1}{\pi} \sqrt{0.25} = 2 \cdot 0.5 = 1.0 \ s$.

$A$ $50 \ cm$ long simple pendulum is suspended from the roof of a cart accelerating in the horizontal direction with constant acceleration $\sqrt{3} \ g \ m/s^2$. The time period of small oscillations of the pendulum about its equilibrium position is $.... \ s$ $\left(g=\pi^2 \ m/s^2\right):-$

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