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Question
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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Solution 1
The bob of the simple pendulum will experience the acceleration due to gravity and the centripetal acceleration provided by the circular motion of the car.
Acceleration due to gravity = g
Centripetal acceleration = `v^2/R`
Where,
v is the uniform speed of the car
R is the radius of the track
Effective acceleration (aeff) is given as:
`a_"eff" = sqrt(g^2 + (v^2/R)^2)`
Time period, `T = 2pi sqrt(1/a_"eff")`
Where, l is the length of the pendulum
:. Time period, `T = 2pi sqrt(1/(g^2 + v^4/R^2))`
Solution 2
In this case, the bob of the pendulum is under the action of two accelerations.
1) Acceleration due to gravity 'g' acting vertically downwards.
2) Centripetal acceleration `a_c = v^2/R` acting along the horizontal direction.
:. Effective acceleration, `g' = sqrt(g^2 + a_c^2)` or `g' = sqrt(g^2 + v^4/R^2)`
Now time period, `T' = 2pi sqrt(1/g) = 2pi sqrt(1/(sqrt(g^2 + v^4/R62)))`
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