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(D) The period of oscillation of a simple pendulum in a stationary lift is given by $T_0 = 2\pi \sqrt{\frac{\ell}{g}}$.When the lift accelerates upwar

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$3g$ upward

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(D) The period of oscillation of a simple pendulum in a stationary lift is given by $T_0 = 2\pi \sqrt{\frac{\ell}{g}}$.When the lift accelerates upward with acceleration $a$,the effective acceleration due to gravity becomes $g_{eff} = g + a$. The new period of oscillation is $T = 2\pi \sqrt{\frac{\ell}{g+a}}$.Given that the new period is $T = \frac{T_0}{2}$,we have:$\frac{T_0}{2} = 2\pi \sqrt{\frac{\ell}{g+a}}$Substituting $T_0 = 2\pi \sqrt{\frac{\ell}{g}}$:$\frac{1}{2} \left( 2\pi \sqrt{\frac{\ell}{g}} \right) = 2\pi \sqrt{\frac{\ell}{g+a}}$$\frac{1}{2} \sqrt{\frac{\ell}{g}} = \sqrt{\frac{\ell}{g+a}}$Squaring both sides:$\frac{1}{4} \cdot \frac{\ell}{g} = \frac{\ell}{g+a}$$\frac{1}{4g} = \frac{1}{g+a}$$g + a = 4g$$a = 3g$Since the effective gravity increased,the acceleration must be directed upward.

$A$ pendulum is suspended in a lift and its period of oscillation when the lift is stationary is $T_0$. What must be the acceleration of the lift for the period of oscillation of the pendulum to be $T_0/2$?

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