Question

A body of mass 12kg is suspended by a coil
of spring of natural length 50 cm and force constant
2x 10^3 N/m. When it is set in to oscillation, what
is the frequency of oscillation.​

Answer 1

Answer:

Given =  mass 12Kg

              length = 50cm  = 0.50m

              spring constant = 2x 10^3 N/m.

to find = frequency of oscillation.​

solution =

apply the hook`s law

F= Kx

mg = kx

x = mg/K

x= 12x9.8 /2x10³

x= 0.5588m

for finding the frequency of oscillation  apply this formula and given values . f = V / λ