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 ?
Text Solution
Verified by Experts
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 with a bob of mass m is suspended from the roof of a car moving with horizontal acceleration a
A simple pendulum with a bob of mass m is suspended from the roof of a car moving with a horizontal acceleration a.
A simple pendulum of length l is suspended in a car that is travelling with a constant speed v around a circle of radius r . If the pendulum undergoes small oscillations about its equilibrium position , what will its freqeuency of oscillation be ?
A simple pendulum of length l and mass m is suspended in a car that is moving with constant speed v around a circle of radius r . Find the period of oscillation and equilibrium position of the pendulum.
A simple pendulum of length l and mass (bob) m is suspended vertically. The string makes an angle theta with the vertical. The restoring force acting on the pendulum is
A simple pendulum of length I and mass (bob) m is suspended vertically. The string makes an angle theta with the vertical. The restoring force acting on the pendulum is
A car of maas M is moving on a horizontal circular path of radius r. At an instant its speed is v and is increasing at a rate a.
A simple pendulum with a bob of mass m is suspended from the roof of a car moving with a horizontal accelertion a. The angle made by the string with verical is
A seconds pendulum is suspended from rof of a vehicle that is moving along a circular track of radius (10)/(sqrt(3))m with speed 10m//s . Its period of oscillation will be (g = 10m//s^(2))