- #1
leroyjenkens
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Homework Statement
A force of 2 lb stretches a spring 1 ft. A 3.2 lb weight is attached to the spring and the system is then immersed in a medium that imparts a damping force numerically equal to 0.4 times the instantaneous velocity. Find the equation of motion if the weight is released from rest 1 ft above the equilibrium position.
Homework Equations
[itex]\frac{d^2t}{dt^2}+\frac{β}{m}\frac{dx}{dt}+\frac{k}{m}x=0[/itex]
The Attempt at a Solution
The initial information says that a force of 2 lb stretches a spring 1 ft. I use that information to get my spring constant. F = ma = ks (s being the distance it stretches).
I use that to get k = 2
Then I use F = ma to get the mass. 3.2 = m*32 (32 [itex]\frac{ft}{s^2}[/itex]=9.8[itex]\frac{m}{s^2}[/itex])
β is given in the problem as 0.4.
So I use all of that and plug it into the formula and get...
[itex]\frac{d^2t}{dt^2}+4\frac{dx}{dt}+20x=0[/itex]
That gives me this... [itex]m^2+4m+20=0[/itex]
I use the quadratic formula to get [itex]-2\pm8i[/itex]
So that gives me [itex]x(t)=e^{-2t}(C_{1}cos8t+C_{2}sin8t)[/itex]
The equation for the velocity is [itex]x'(t)=-2(-8C_{1}sin8t+8C_{2}cos8t)[/itex]
According to the problem, the initial position is -1 ft because it is released 1 foot above the equilibrium position, and down is positive. The initial velocity is 0 since it starts from rest.
But when I solve for x(0), I get [itex]C_{1}=-1[/itex] and when I solve for x'(0), I get [itex]C_{2}=0[/itex]
The back of the book says those are the wrong constants.
And it has a different argument in the trig functions.
Here's the answer in the back of the book: [itex]x(t)=e^{-2t}(-cos4t-\frac{1}{2}sin4t)[/itex]