Question

A body dropped from a height h onto the floor makes elastic collision with the floor. The frequency of oscillation of its periodic motion is

Answer 1

$\frac{1}{2 \sqrt{2} }$
frequency equals to 1 divided by time period mark me as a brainiest answer if you like it

Answer 2

Hey dear,

◆ Answer-
f = √(g/8h)

◆ Explaination-
Using Newton's 2nd kinematic eqn,
s = ut + 1/2 at^2

Here, s = h, u = 0, & a = g, hence-
h = 1/2 gt^2
t = √(2h/g)

In elastic collision, time to go up is same as that to go down. t1 = t2 = t

Period of oscillation will be -
T = time to go down + time to go up
T = t + t = 2t
T = 2 √(2h/g)

Frequency of oscillation is reciprocal of period.
f = 1/T
f = √(g/8h)

Hope this helps...