Question

obtained and expression for period of simple pendulum performing SHM​

Answer 1

Answer:

time period of the oscillating SHM

$T = 2\pi \sqrt{l/g}$

or

$T = 2\pi \sqrt{m/k}$

or

$T = 2\pi \sqrt{mu/k}$

where mu is the reduced mass

mu  =  m1/m1+ m2