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13.Oscillations
easy
There is a simple pendulum hanging from the ceiling of a lift. When the lift is stand still, the time period of the pendulum is $T$. If the resultant acceleration becomes $g/4,$ then the new time period of the pendulum is
A
$0.8 \,T$
B
$0.25\, T$
C
$2\, T$
D
$4\, T$
Solution
(c) When lift is at rest, $T = 2\pi \sqrt {l/g} $
If acceleration becomes $g/4$ then
$T' = 2\pi \sqrt {\frac{l}{{g/4}}} = 2\pi \sqrt {\frac{{4l}}{g}} = 2 \times T$
Standard 11
Physics