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(A) The period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{L}{g_{eff}}}$,where $g_{eff}$ is the effective acceleration due to gravity.When

Vedclass · Accepted Answer

$2 \pi \sqrt{L}(a^{2}+g^{2})^{-\frac{1}{4}}$

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(A) The period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{L}{g_{eff}}}$,where $g_{eff}$ is the effective acceleration due to gravity.When the trolley moves horizontally with acceleration $a$,the effective acceleration $g_{eff}$ is the vector sum of the gravitational acceleration $g$ (acting downwards) and the pseudo-acceleration $a$ (acting horizontally in the opposite direction).Since these two accelerations are at right angles to each other,the magnitude of the effective acceleration is $g_{eff} = \sqrt{g^2 + a^2} = (a^2 + g^2)^{\frac{1}{2}}$.Substituting this into the formula for the time period:$T = 2 \pi \sqrt{\frac{L}{(a^2 + g^2)^{\frac{1}{2}}}} = 2 \pi \sqrt{L} \cdot (a^2 + g^2)^{-\frac{1}{4}}$.

$A$ simple pendulum of length $L$ is suspended from the roof of a trolley. The trolley moves in a horizontal direction with an acceleration $a$. What would be the period of oscillation of the simple pendulum? [$g$ is the acceleration due to gravity]

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