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)
a) Find the equation which applies to this
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: 2D2
+ 98D = 0. It's roots are 7i and -7i .Determine
xh: using roots 7i and
-7i of the characteristic polynomial, we obtain
Our first guess for xp should be: f(t) = A1sin(7t)+A2cos(7t) BUT, since this is similar to the term C2sin(7t) in the equation of xh, we must multiply xp by t, therefore we guess
using our initial equation (2x" + 98x = -196sin(7t))
substitute xp in for x:
Therefore: A1 = 7 and A2
= 0. Substituting these values into the equation
xp = A1tsin(7t) + A2tcos(7t)
Now solve for C1 and C2:
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/s2.
We have mg=ks , so k=mg/s = 1(9.8)/.098 = 100kg/s2
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(C1/C2) and A = sqrt( C12 + C2 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(10t-1.520837931) , 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.