Spring SHM System ```Name: Brian S. Status: educator Age: 30s Location: N/A Country: N/A Date: 2/26/2003 ``` Question: I need some help with the energy equation for the SHM f a vertical spring. I would like to know what the equation would look like for the max ocity, velocity at equilibrium, of a mass hanging from a vertical spring. The COE equation gives us this DKE + DPEgravity + DPEspring = 0 We cannot cancel PE gravity because if you pull down on a mass and let it go, as the mass reaches equilibrium, the velocity should be a maximum. Is this correct? Vmax = SQRT(k/m*X^2 - 2gh) where h = X the displacement of the spring.... I am also wondering if I need to take in account the energy the spring has when the mass reaches equilibrium? Anyway, I just need to know is this equation for V at equilibrium for a mass hanging from a vertical spring is correct? Replies: Yes, if h=X is the displacement of the mass downward from its equilibrium position. In principle, you should take the energy of the spring into account, but it's better to start students off with an idealized, massless spring, and anyway, it is usually a pretty good approximation. Tim Mooney This is a rather confusing problem partly because gravity can be completely ignored if we measure displacements from the equilibrium position of the mass hanging on the spring. Notice that the force acting on the mass is F = -ky - mg if y is measured upwards from the position of the mass when the spring is not extended and so the spring is exerting no force on the mass. k is the spring constant and mg is the weight of the mass. If you now hang the mass on the spring, it will come to an equilibrium when F = 0 which happens when y = -mg/k. Right? Now do a change of variable to z = y + d, where d = mg/k, so z is measured from the mass position when it is hanging on the spring at rest. Then F = -ky - mg becomes F = -k(z-d) - mg = -kz (use d=mg/k) and we can ignore gravity. In the energy description, which you used, we can show the same thing, though it needs a little more algebra. If the mass is pulled down to y = -d -h and released from rest, its speed at y = -d, which is the equilibrium position is v = sqrt((k/m)*h^2). Using z the mass is pulled down to z = -h. Then its speed at z = 0 is the same. E = mv^2/2 + mgy + ky^2/2 = mv^2/2 + kz^2/2 - kd^2/2 (using d=mg/k) Since kd^2/2 is a constant, it has no effect. Gravity has disappeared! I recommend you draw a careful diagram and go through the algebra carefully. If it does not work out and you want to see it, send me an email directly via NEWTON (SYSOP@newton.dep.anl.gov). Best, Dick Plano, Professor of Physics emeritus, Rutgers University So if you measure the displacement of the mass from its equilibrium position, the effect of gravity vanishes! What really happens, of course, is that the extra spring tension just exactly balances the weight of the mass. Brian, The first factor you seem to have missed is the initial stretch factor as it relates to the equilibrium position. Initial stretch=X, where zero stretch is zero spring force and zero spring potential energy. Let initial position be zero height, zero gravitational potential energy Initial speed=0, initial KE=0. Final position,x,is at equilibrium: kx=mg, x=mg/k. Note that x < X. v=final velocity final height=X-x, initial stretch-final stretch ```{ (1/2)mv^2-0 } + { mg(X-x)-0 } + { (1/2)kx^2-(1/2)kX^2 } = 0 (1/2)mv^2 = -mg(X-x) - (1/2)kx^2 + (1/2)kX^2 v^2 = -2g(X-x) - (k/m)x^2 + (k/m)X^2 v^2 = 2gx - 2gX - (k/m)x^2 + (k/m)X^2, substitute x=mg/k v^2 = 2mg^2/k - 2gX - mg^2/k + (k/m)X^2 v^2 = mg^2/k + (k/m)X^2 - 2gX ``` Dr. Ken Mellendorf Physics Instructor Illinois Central College Click here to return to the Physics Archives

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