A simple pendulum of length `1m` is allowed to oscillate with amplitude `2^(@)`. it collides at `T` to the vertical its time period will be (use `g = pi^(2)`)
A
`2//3 sec`
B
`4//3 sec`
C
`2 sec`
D
None of these
Text Solution
Verified by Experts
The correct Answer is:
B
Time period for half part : `T = 2pi sqrt((l)/(g)) = 2pisqrt((1)/(g)) = (2pi)/(pi) = 2` sec. So `2^(@)` part will be covered in a time `t = (T)/(2) = 1 sec`. For the left `1^(@)` part : `theta = theta_(0) sin(omegat) rArr 1^(@) = 2^(@)sin((2pi)/(T) xxt) rArr (1)/(2) = sin ((2pi)/(2) xx t) rArr (pi)/(6) = pi xx t rArr t = 1//6 sec` Total time `= T/2 + 2t rArr 1 + 2 xx (1)/(6) = 1 + 1/3 = 4/3 ` sec.
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