Recent content by s.g.g

sorry tim, disregard that rubbish last remark. it is completely incorrect. i say that x-y=p & y-z=q i then find that d2p/dt2=Bcos(omega*t) +Aq & d2q/dt2=A(p-2q) But same as before, i cannot solve for p and q, what am i missing?

Ha sorry:smile:, Heres what iv done. say y-x=p & z-y=q hence d2x/dt2 = d2p/dt2+d2y/dt2, d2y/dt2 = d2p/dt2+d2x/dt2 = d2q/dt2 + d2z/dt2 & d2z/dt2= d2q/dt2+d2y/dt2 the original equations then become d2p/dt2=A(2p-q-Bcos(omega*t)) d2q/dt2=A(2q-p) but I am then stuck with the same...

I still can't do it, I am trying to compute d2(x-y) /dt2 and letting (x-y) =gamma. But i cannot solve for gamma

Ha I'm starting to think the answer is so easy i can't see it. That equals the LHS of the sum of the equations. So therefor d2/dt2(x+y+z) = AB*cos (omega*t) after cancelling, integrating twice with respect to t gives x(t)+y(t)+z(t) = -(AB/omega2)cos(omega*t) + C How do i...

Sorry, yeah the last equation should be d^2z/dt2 = A(y-z) x, y and z are all functions of time. d^2(x+y+z)/dt^2 = d^2x/dt^2 + d^2y/dt^2 + d^2z/dt^2 isn't it?

Homework Statement I have 3 masses in 1-D connected by two springs. A driving force is exerted on the first mass and i need to derive the equation of motion of the last mass. I have worked out the Lagrangian to determine the equations of motion but cannot solve for z. Homework Equations The...

i was uncertain whether i could ascribe an equilibrium position to each mass as it is determined by its position relative to the adjacent masses rather than a fixed point in space. Hence it is constantly changing. that is why i thought i had to introduce the value J. Is this wrong?

Homework Statement A system consists of 3 identical masses (A,B & C) in a line, connected by 2 springs of spring constant k. Motion is restricted to 1 dimension. at t=0 the masses are at rest. Mass A is the subjected to a driving force given by: F=F0*cos(omega*t) Calculate the...