Recent content by Chip90

right my equation is? P0*g*l3*y'' - y*pw*g*l2 + p0*g*l3=0 this has the solution y(t)= A * cos (w*t + \phi) so how do i figure out \phi I know \omega from F=Fb which gives me \omega= sqrt (pW*g/pO*l)

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?

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.

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

Homework Statement A cork with a density \rho0 in the form of a cube of side length l floats on water with a density of \rhow. The pressure in water depends on depth h from the surface as P=\rhow *g*h. A. Find the equilibrium depth of the bottom surface of the cube (how much length is below...

ok taking the integral of that yields Q\int\frac{dE}{dn} = -2\pi \int E and the integral is from 0 to 100 because you want to get the energy after 100 oscillations right? but Ef is a fixed value, so how do I take an integral over this? I think it should go to this: Q Efinal \int\frac{1}{dn}...

I set it up as Q= 2* \pi \int0.003425/0.000845 dn from 0 to 100 so I think I forgot to numltiply by 100, so then I would get 2,545.. that doesn't seem right

ok so I should have E= 0.003425 Joules and the integral should be from 0 to 100, which would give me a Q factor of 25.45. Then use Q=\omega 0/\gamma to give me \gamma=0.033

Ok, so is this an exponential decay? The E=E0*e-y*t and the energy level is 1/100 so 1/100=e-y*t. But then I don't know the time... or do i want the situation where the final energy is given my the velocity of the weight? In which case Efinal = 0.5 * m * v2 Where Efinal=0.000845 Joules? and I...

Right I meant omega naut so that is right that was supposed to be pi so ω= 2*\pi*f Well Q is a dimensionless quantity, it represent the ratio of Energy stored to Energy lost. I also know that for large values of Q, Q= about number of oscillations. Q= 2*\pi*f*\stackrel{Energy...

well Q= 2*pi*f/y and Q=w1/y=w1/y and w= 2*m*f But how do 100 oscillations factor into all this?

Homework Statement A 100g mass is suspended on a rubber band with a k coefficient of 2.74 N/m. The original amplitude of the oscillations is 5cm and after 100 oscillations, the maximum speed of the weight is 0.13 m/s. Find the damping coefficient y. Homework Equations d2x/dt2 + γdx/dt...

is your calculator in the correct mode?

ok so to find the inner B field, I use which would lead me to say B=Uo*I/ 2*pi*r and for the outer B=Uo*I/L

Homework Statement There is a wire with current I ,charge density j, and radius R. Find the Magnetic Field outside the wire (a>r) and inside the wire (a<r). Homework Equations Biot- Savart or Amperes law I=JdotA The Attempt at a Solution Not sure how to attempt it or which...