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.

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

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Thiên Di 4 weeks 2021-08-17T12:11:57+00:00 1 Answers 1 views 0

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    2021-08-17T12:13:52+00:00

    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

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