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A pendulum has a period T for small osil...

A pendulum has a period `T` for small osillations. An obstacle is placed directly beneath the pivot, so that only the lowest one - quarter of the string can follow the pendulum bob when it swings to the left of its resting position. The pendulum is released from rest at a certain point. How long will it take to return to that point again ? In answering this question, you may assume that the angle between the moving string and the vertical stays small throughout the motion.

Text Solution

Verified by Experts

The correct Answer is:
B
Half the oscillation is completed with length `l` and rest half with length `(l)/(4)`.
`T = (T_(1))/(2) + (T_(2))/(2) = (1)/(2)[2pisqrt((l)/(g)) + 2pisqrt((l)/((4)/(g)))]`
`= (3)/(4)[2pi sqrt((l)/(g))] = (3)/(4) T`, where `T = 2pi sqrt((l)/(g)`
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Knowledge Check

  • A pendulum of length L swings from rest to rest n times in one second. The value of acceleration due to gravity is

    A
    `4pi^(2)n^(2)L`
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  • A pendulum has time period T for small oscillations. An obstacle P is situated below the point of suspension O at a distance (3l)/4 . The pendulum is released from rest. Throughout the motion, the moving string makes small angle with vertical. Time after which the pendulum returns back to its initial position is

    A
    `T`
    B
    `(3T)/(4)`
    C
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  • A simple pendulum consisting of a mass M attached to a string of length L is released from rest at an angle alpha . A pin is located at a distance l below the pivot point. When the pendulum swings down, the string hits the pin as shown in figure. The maximum angle theta which the string makes with the vertical after hitting the pin is

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    D
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