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A simple pendulum of length 1m  is freely suspended from the ceiling of an elevator. The time period  of small oscillations as the elevator moves up with an acceleration of 2m/s² is (use g=10 m/s ²)

Answer:

Explanation:

 Given,

 Length of simple pendulum (L) =1m

 Acceleration (a) =2 m/s²

 Gravitation  acceleration (g)=10 m/s²

 We know that,

    $T= 2 \pi \sqrt{\frac{l}{g+a}}=2\pi\sqrt{\frac{1}{10+2}}$

$= 2 \pi \sqrt{\frac{1}{12}}=\pi\sqrt{\frac{4}{12}}$

$I= \frac{\pi}{\sqrt{3}}s$