Answer:
f = 5.3 Hz
Explanation:
To solve this problem, let's find the equation that describes the process, using Newton's second law
∑ F = ma
where the acceleration is
a = [tex]\frac{d^2 y}{dt^2 }[/tex]
B- W = m \frac{d^2 y}{dt^2 }
To solve this problem we create a change in the reference system, we place the zero at the equilibrium point
B = W
In this frame of reference, the variable y' when it is oscillating is positive and negative, therefore Newton's equation remains
B’= m [tex]\frac{d^2 y'}{dt^2 }[/tex]
the thrust is given by the Archimedes relation
B = ρ_liquid g V_liquid
the volume is
V = π r² y'
we substitute
- ρ_liquid g π r² y’ = m \frac{d^2 y'}{dt^2 }
[tex]\frac{d^2 y'}{dt^2} + \rho_liquid \ g \ \pi r^2/m ) y' \ =0[/tex]
this differential equation has a solution of type
y = A cos (wt + Ф)
where
w² = ρ_liquid g π r² /m
angular velocity and frequency are related
w = 2π f
we substitute
4π² f² = ρ_liquid g π r² / m
f = [tex]\frac{1}{2\pi } \ \sqrt{ \frac{ \rho_{liquid} \ \pi r^2 \ g}{m } }[/tex]
calculate
f = [tex]\frac{1}{2 \pi } \sqrt{ \frac{ 1000 \ \pi \ 0.03^2 \ 9.8 }{0.025} }[/tex]
f = 5.3 Hz
