i need to know how to do 2, 4 & 6 (highlighted) A spring with a 3-kg mass is hel

Question

i need to know how to do 2, 4 & 6 (highlighted)

A spring with a 3-kg mass is held stretched 0.6 m beyond its natural length by a force of 20 N. If the spring begins at its equilibrium position but a push gives it an initial velocity of 1.2 m/s, find the position of the mass after t seconds. A spring with a 4-kg mass has natural length 1 m and is maintained stretched to a length of 1.3 m by a force of 24.3 N. If the spring is compressed to a length of 0.8 m and then released with zero velocity, find the position of the mass at any time t. A spring with a mass of 2 kg has damping constant 14, and a force of 6 N is required to keep the spring stretched 0.5 m beyond its natural length. The spring is stretched 1 m beyond its natural length and then released with zero velocity. Find the position of the mass at any time t. A spring with a mass of 3 kg has damping constant 30 and spring constant 123. Find the position of the mass at time t if it starts at the equilibrium position with a velocity of 2 m/s. Graph the position function of the mass. For the spring in Exercise 3, find the mass that would produce critical damping. For the spring in Exercise 4, find the damping constant that would produce critical damping. A spring has a mass of 1 kg and its spring constant is k = 100. The spring is released at a point 0.1 m above its equilibrium position. Graph the position function for the following values of the damping constant c: 10, 15, 20, 25, 30. What type of damping occurs in each case? A spring has a mass of 1 kg and its damping constant is c = 10. The spring starts from its equilibrium position with a velocity of 1 m/s. Graph the position function for the following

Explanation / Answer

4. simple harmonic motion with damping.

balancing the forces: mx'' + 2m gamma x' + kx = 0

this is a second order homogenous differential equation with solution x=exp (lambda t) where lambda can be found by substitution of x=exp (lambda t) into the equation and solving the quadratic auxillary equaton.

so u get something like mlambda^2 + 2m gamma lambda + k =0

solving you get lambda = (1/2m) (-2mgamma +/- sqrt ((2mgamma)^2 - 4mk))

putting in the numbers = (1/(2*3)*(-2*3*30 -sqrt((2*3*30)^2-4*3*123))= -59.3 (here we take the solution that gives us negative because we want the damped solution)

x=Aexp(-59.3t + phi)

x'=-59.3A exp (-59.3 t + phi)

put in the initial conditions (x'=2 and x=0) : phi =0, A = 29.65

the graphing part i'll just let you do it yourself, ain't that hard (jsut a decaying exponential).

6. Critical damping occurs at 4m^2 gamma - 4mk = 0 (basically the stuff in the sqroot sign becomes 0).

so k = m.gamma = 3*30 = 90

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