Question

The pendulum of length 'L' swing from
mean postion to mean 'N' times in one
second. the value of acceleration due to
gravity is .
(a)
na
(b)
21t?n
(c) (r'n L)/2
(d) 41?n?​

Answer 1

Answer:

$\frac{1}{N} =2\pi\sqrt{\frac{L}{g} } \\\sqrt{g} = 2\pi \times N \sqrt{L}\\g=4 {\pi}^{ 2}N^2L$

Explanation:

T  = $2\pi\sqrt{\frac{L}{g} }$