A simple pendulum of length l is suspen...

A simple pendulum of length `l` is suspended from the celing of a cart which is sliding without friction on as inclined plane of inclination theta . What will be the time period of the pendulum?

Text Solution

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The correct Answer is:

A, B, C

Here, point of suspension has an acceleration. `a = g sin theta` (down the plane). Further, `g` can be resolved into two components `g sin theta` (along the plane) and `g cos theta` (perpendicular to plane).
`:. g_(eff) = g - a = g + ( -a)`
Resultant of g and `-a` will be `g cos theta`.
`:. g_(eff) = g cos theta`
(perpendicular to plane)
`:. T = 2pi sqrt((l)/(|g_(eff)|)) = 2pi sqrt ((l)/(g cos theta)`

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The period of oscillation of a simple pendulum of length l suspended from the roof of a vehicle which moves down without friction on an inclined plane of inclination theta , is given by

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`2pi sqrt((l)/(g cos alpha))`

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