Applications of ODE; damped motion

[leroyjenkens]

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]

 

[LCKurtz]

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]

I would check that last step, especially since the answer has 4t in its arguments for the trig functions.

 

[leroyjenkens]

I would check that last step, especially since the answer has 4t in its arguments for the trig functions.

Ah, you got me. I always divide the left side of the numerator by the denominator, but forget to divide the right side by the denominator as well. Thanks. I'll check that to see if it fixes my problems.

Ok I'm still getting 0 for [itex]C_2[/itex]

My [itex]C_1[/itex] is correct. That equals -1.

Am I wrong that x'(0) is not equal to 0? That's the only way I can see getting [itex]C_2 = \frac{-1}{2}[/itex]

 

[vela]

What happened to the exponential factor when you differentiated to find x'(t)?

 

[leroyjenkens]

What happened to the exponential factor when you differentiated to find x'(t)?

I messed up, it's supposed to be there in x'(t), but it turns to 1 in x'(0), which leaves -2 outside the parentheses to turn my 4 into -8. But I still have that 0 on the left side, so I don't see how I could get the -1/2 that the book has for [itex]C_2[/itex].
What do you think?

 

[vela]

You didn't differentiate correctly if you didn't use the product rule, which I don't think you're doing.