An object weighing 34 lbs. is attached to a spring with spring constant 1 lb/ft. This spring-mass system is subject to a damping forece numerically equal to twice the velocity. At time t=0 the object is lowered 3 inches below its equilibrium position and given an initial velocity of 1 ft./sec in the upward direction.(adsbygoogle = window.adsbygoogle || []).push({});

Show that the object will pass through it equilibrium position exactly once and then creep back towards its equilibrium position.

Thank you

A Response by Kevin:

2004-04-21 at 13:41GMT

This is what I have so far. Please help

So the first thing we need to do is formulate the differential equation. Based on one of Newton's Laws of Motion (forgot which one offhand), the sum of the forces on a body equals the product of the mass and acceleration of the body. If we denote the displacement (in feet) of the mass from its equilibrium position by x (with positive x corresponding to the mass lying below its equilibrium position), then

spring force = -k*x, k = 1 lb/ft (Hook's Law)

damping force = -c*(dx/dt), c = 2 lb/(ft/s)

Since acceleration is the second time derivative of displacement, we arrive at

m*(d^2x/dt^2) = -k*x - c*(dx/dt)

Letting ' denote differentiation by t, we obtain the differential equation

m*x'' + c*x' + k*x = 0

with initial conditions

x(0) = x0 = 3 inches = 1/4 ft

x'(0) = v0 = -1 ft/s

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# Spring problem stuck

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