A simple pendulum is suspended from the ceili | vedclass

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(B) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g_{eff}}}$.When the lift is at rest,$g_{eff} = g$,so $T = 2\pi \sqrt{\fr

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$3g$

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(B) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g_{eff}}}$.When the lift is at rest,$g_{eff} = g$,so $T = 2\pi \sqrt{\frac{L}{g}}$.When the lift accelerates upward with acceleration $a$,the effective gravity becomes $g_{eff} = g + a$.The new time period is $T' = 2\pi \sqrt{\frac{L}{g+a}}$.Given that $T' = \frac{T}{2}$,we have $\frac{T}{2} = 2\pi \sqrt{\frac{L}{g+a}}$.Substituting $T = 2\pi \sqrt{\frac{L}{g}}$,we get $\frac{1}{2} \left( 2\pi \sqrt{\frac{L}{g}} \right) = 2\pi \sqrt{\frac{L}{g+a}}$.Squaring both sides: $\frac{1}{4} \left( \frac{L}{g} \right) = \frac{L}{g+a}$.This simplifies to $g + a = 4g$,which gives $a = 3g$.

$A$ simple pendulum is suspended from the ceiling of a lift. When the lift is at rest,its period is $T$. With what acceleration $a$ should the lift be accelerated upward in order to reduce the period to $\frac{T}{2}$? (Take $g$ as the acceleration due to gravity.)

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