A simple pendulum of length L and having a bob of mass m is suspended in a car. The car is moving on a circular trrack of radius R with a uniform speed v. If the pendulum makes small oscillations in a radial direction about its equilibrium position, its time period of oscillation is
Text Solution
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Centripetal acceleration `a_(c) = (v^(2))/(R)` & Acceleration due to gravity `= g` So `g_(eff) = sqrt(g^(2) + ((v^(2))/(R))^(2)) rArr "Time period" T = 2pisqrt((L)/(g_(eff))) = 2pisqrt((L)/(sqrt(g^(2)+(v^(4))/(R^(2)))))`
A simple pendulum of length l and having a bob of mass M is suspended ina car. The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small oscillations in a radial direction about its equilibrium, what will be its time period ?
A simple pendulum of length l and having a bob of mass M is suspended ina car. The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small oscillations in a radial direction about its equilibrium, what will be its time period ?
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