
#1
Apr1811, 12:33 PM

P: 55

1. The problem statement, all variables and given/known data
A cork with a density [tex]\rho[/tex]_{0} in the form of a cube of side length l floats on water with a density of [tex]\rho[/tex]_{w}. The pressure in water depends on depth h from the surface as P=[tex]\rho[/tex]_{w} *g*h. A. Find the equilibrium depth of the bottom surface of the cube (how much length is below the surface) Let the equilibrium position of the bottom surface be at y=0. At time t=o, the position of the bottom surface (below the equilibrium point) is at y=A_{0} and the cube is moving upwards with a speed of V_{0}. B. Find the behavior of y(t) C. Find the earliest time it take to reach the origin and the next time it takes to reach it again. Find the maximum speed the cork can have. D. Friction is now present with friction force aV(y,t) where a is a contant and V(y,t) is the velocity. Assume the friction results in a very lightly damped motion. Find the Q value. 2. Relevant equations All in next part. 3. The attempt at a solution This problem should be easy enough but I am having some problems. A. The depth under water is [tex]\frac{P0}{Pw}[/tex] *l B. I did the force diagram on the block, with F_{g} in the negative y direction and F_{b} in the positive y direction. F=ma Fb Fg = ma where Fb=[tex]\rho[/tex]w*g*y*l^{2} and Fg=m*g=[tex]\rho[/tex]o*g*l^{3} this gives me [tex]\rho[/tex]w*g*l^{2}*y  m*g=[tex]\rho[/tex]o*g*l^{3} = m*a where m= ([tex]\rho[/tex]w+[tex]\rho[/tex])o*l^{3} a= [tex]\frac{d2*x}{dt2}[/tex] [tex]\rho[/tex]w*g*l^{2}*y  m*g=[tex]\rho[/tex]o*g*l^{3} = ([tex]\rho[/tex]w+[tex]\rho[/tex]o)*l^{3}* [tex]\frac{d2*x}{dt2}[/tex] is this the right setup? my prof. said there was something wrong with this, but I have no idea what.. I am also not sure how to approach part B. The undamped, unforced harmonic osc. eq. is [tex]\frac{d2*x}{dt2}[/tex]+ w_{0}^{2}*x=0 



#2
Apr1811, 07:50 PM

P: 55

any ideas why that setup would be wrong? I am still not seeing the problem with it..




#3
Apr1811, 08:05 PM

P: 242

the m in m*a should be just rho_o*g*l^3, not (rho_w+rho_o)*g*l^3




#4
Apr1811, 08:06 PM

P: 55

Buoyancy and Harmonic Oscillation
i see and what is the reason for that? does that still include for the fact that as the block is moving down, water is moving away and v.v.




#5
Apr1811, 08:14 PM

P: 242

you are summing the forces on the cork, so the m should just be the mass of the cork. the direction the block is moving does not matter in setting up the differential equation, just in the initial conditions.




#6
Apr1811, 08:22 PM

P: 55

ah, that makes sense. now that i have this equation with something equaling m* a
do i just take the second integral of both sides to get the y(t) function? 



#7
Apr1811, 08:32 PM

P: 242

no, not so simple. in fact, now that i look closely, you are using x and y to mean the same thing in your equations. your equation is of the form: ay'' +by +c = 0. characteristic equation should get you the solution.




#8
Apr1811, 08:53 PM

P: 55

right my equation is? P_{0}*g*l^{3}*y''  y*p_{w}*g*l^{2} + p_{0}*g*l^{3}=0
this has the solution y(t)= A * cos (w*t + [tex]\phi[/tex]) so how do i figure out [tex]\phi[/tex] I know [tex]\omega[/tex] from F=Fb which gives me [tex]\omega[/tex]= sqrt (pW*g/pO*l) 


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