<p>A 128 pound weight is attached to the lower end of a coiled spring suspended
ID: 1944763 • Letter: #
Question
<p>A 128 pound weight is attached to the lower end of a coiled spring suspended from the ceiling. The weight comes to rest at its equilibrium position, thereby stretching the spring 8 feet. At time t = 0, the weight is positioned at √3 feet below the equilibrium and given a downward velocity of 2 feet per second. </p><p>Determine:</p>
<p>a) The equation of motion of the weight as a function of time. </p>
Explanation / Answer
Let m be the mass of the weight, y0 the equillibrium position, y the displacement from y0, k the spring constant, and g the acceleration due to gravity. The mass m, sees a net force of: ma = mg + k(y0-y) but a = d^2y/dt^2 = y" so my" = mg + k(y0-y) ---> my" + ky = mg - ky0 = 0 since mg and ky0 balance (defintion of equillibrium) Note: k= mg/y0 = 48 lbs/8 ft = 6 lbs/ft Hence my" + ky = 0 --> y" + k/m y = 0 define w^2 = k/m, then y = Acos(w*t) + Bsin(w*t) w^2 = 6 lbs /f /(48lbs/32 f/s^2) = 4 rad^2/s^2 ---> w = 2 rad/s ---> frequency f = w/(2*pi) = 0.637 Hz Now apply initial conditions: At t=0, y(0) = 1 ft = Y, and dy/dt = y' (0) = 2 ft/s =V y(0) = Y = A y'(0) = V = wB ---> B = V/w = 1 ft So y(t) = cos(2t) + sin(2t) Period is 1/frequncy = 1.57 sec You can use the equation of motion to do the rest.
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