A simple pendulum of length l and mass m of t | vedclass

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Question

(C) The bob of the pendulum experiences two accelerations in the frame of the car: the acceleration due to gravity $g$ acting downwards and the centri

Vedclass · Accepted Answer

$\frac{1}{2\pi} \sqrt{\frac{\sqrt{g^2 + \frac{v^4}{R^2}}}{l}}$

Vedclass · Answer

(C) The bob of the pendulum experiences two accelerations in the frame of the car: the acceleration due to gravity $g$ acting downwards and the centrifugal acceleration $a_c = \frac{v^2}{R}$ acting horizontally outwards.The effective acceleration $g_{eff}$ is the vector sum of these two perpendicular accelerations:$g_{eff} = \sqrt{g^2 + a_c^2} = \sqrt{g^2 + \left(\frac{v^2}{R}\right)^2} = \sqrt{g^2 + \frac{v^4}{R^2}}$The time period $T$ of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g_{eff}}}$.Substituting the value of $g_{eff}$:$T = 2\pi \sqrt{\frac{l}{\sqrt{g^2 + \frac{v^4}{R^2}}}}$The frequency $f$ is the reciprocal of the time period:$f = \frac{1}{T} = \frac{1}{2\pi} \sqrt{\frac{\sqrt{g^2 + \frac{v^4}{R^2}}}{l}}$Thus,the correct option is $C$.

$A$ simple pendulum of length $l$ and mass $m$ of the bob is suspended in a car that is travelling with a constant speed $v$ around a circular path of radius $R$. If the pendulum undergoes oscillations with small amplitude about its equilibrium position,the frequency of its oscillations will be

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