Question

If a body of mass m attached to a spring
oscillates in a non-viscous liquid of
density d then its time period will be
$2\pi \sqrt{m(1 - d \div p) \div k}$
where d is the density of non viscous liquid
and p is the density of body
​How?...please derive it...

Answer 1

Answer:

Explanation:

The time period of oscillation T of a small liquid under surface tension sigma depends upon the density rho the radius r and sigma. Using dimensional analysis show that The is directly proportional to root of rho r cube by sigma