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 track of radius R with a uniform speed v. If the pendulum makes small oscillations in a radial direction about its equilibrium position, what will be its time period?
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Hii dear,
◆ Answer-
T = 2π√l(g^2+v^4/r^2)^(-1/4)
◆ Solution-
Consider, a
Here, the pendulum will experience 2 acceleration-
a) Gravitational acceleration a1 = g
b) Centripetal acceleration a2 = v^2/r
Resultant acceleration-
g' = √(g^2+v^4/r^2)
Period of oscillation of pendulum is
T = 2π √(l/g')
T = 2π√l(g^2+v^4/r^2)^(-1/4)
Hope this helps you...
◆ Answer-
T = 2π√l(g^2+v^4/r^2)^(-1/4)
◆ Solution-
Consider, a
Here, the pendulum will experience 2 acceleration-
a) Gravitational acceleration a1 = g
b) Centripetal acceleration a2 = v^2/r
Resultant acceleration-
g' = √(g^2+v^4/r^2)
Period of oscillation of pendulum is
T = 2π √(l/g')
T = 2π√l(g^2+v^4/r^2)^(-1/4)
Hope this helps you...
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