Question

A circular cylinder has a diameter of 3.0 cm and a mass of 25 g. It floats in water with its long axis perpendicular to the water's surface. It is pushed down into the water by a small distance and released; it then bobs up and down. Part A What is the oscillation frequency

Answer 1

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 = $\frac{d^2 y}{dt^2 }$

        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 $\frac{d^2 y'}{dt^2 }$

           

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 }

          $\frac{d^2 y'}{dt^2} + \rho_liquid \ g \ \pi r^2/m ) y' \ =0$

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 = $\frac{1}{2\pi } \ \sqrt{ \frac{ \rho_{liquid} \ \pi r^2 \ g}{m } }$

calculate

         f = $\frac{1}{2 \pi } \sqrt{ \frac{ 1000 \ \pi \ 0.03^2 \ 9.8 }{0.025} }$

         f = 5.3 Hz