Prof. D. Joyner
1. A mass of 2 kg is attached to a spring having
spring constant 98 N/m with negligible damping. The mass starts at equilibrium
with initial velocity of 7m/s downward, and an external force of 196sin(7t)
acts.
a) Find the equation which applies to this
case:
mx" + kx = 196sin(7t) (m=2kg , k=98N/m) , so
2x" + 98x = 196sin(7t)
Determine the characteristic polynomial and
solve for the roots:
The characteristic polynomial is: 2D^{2}
+ 98D = 0. It's roots are 7i and 7i .Determine
x_{h}: using roots 7i and
7i of the characteristic polynomial, we obtain
Our first guess for x_{p }should be: f(t) = A_{1}sin(7t)+A_{2}cos(7t) BUT, since this is similar to the term C_{2}sin(7t) in the equation of x_{h}, we must multiply x_{p} by t, therefore we guess
using our initial equation (2x" + 98x = 196sin(7t))
substitute x_{p} in for x:






Therefore: A_{1} = 7 and A_{2}
= 0. Substituting these values into the equation
x_{p} = A_{1}tsin(7t) + A_{2}tcos(7t)
gives:
Now solve for C_{1} and C_{2}:
since:
b) sketch x(t), 0<t<1:
For pure resonance: r=sqrt(k/m)
We have k=.5 and m=1/32 .
3. A spring is suspended from the ceiling.
The spring was stretched .098m after a 1 kg mass was attached to it and
has negligible damping. Take g=9.8m/s^{2}.
We have mg=ks , so k=mg/s = 1(9.8)/.098 = 100kg/s^{2}
.
a) Find the displacement if the mass
is initially released with a 1m/s downward velocity 2m up from equilibrium.
We must solve mx" + kx = 0 or, in our case, x" + 100x = 0 . In general,
b)Write the displacement in the form Asin(w
t + f).
Formulas: f = tan^{1}(C_{1}/C_{2}) and A = sqrt( C_{1}^{2} + C_{2 }^{2})
We know w = 10, f = tan^{1}(2) = 1.107rad , so
c) Find the first local extrema of x(t), t>0,
and indicate if it is a maximum or a minimum of x(t).
We solve x'(t) = 0 = sqrt(401)*cos(10t1.520837931) , so
Next, we test for maximum or minimum: x"(t) =
10sqrt(401)sin(10t  1.520837931) , so x"(.3091634258) = 200.2498440.
=> .3091634258 sec is on the maximum of the curve and therefore the minimum of the springs actual displacement.