Answers to Team quiz, SM212, 9-25-98

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: 2D2 + 98D = 0. It's roots are 7i and -7i .Determine xh:  using roots 7i and -7i of the characteristic polynomial, we obtain

xh=C1cos(7t) + C2sin(7t)

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

xp=linear combination of f(x), f'(x)
xp = A1tcos(7t) + A2tsin(7t)         (ignore sin(7t), cos(7t) terms since they occur in xh)

using our initial equation (2x" + 98x = -196sin(7t)) substitute xp in for x:

2x" = 2 xp
2(-49A1tcos(7t) + 14A2 cos(7t)  - 49A2tsin(7t) -14 A1sin(7t)) 
98x = 98x
98(A1tcos(7t) + A2tsin(7t) ) 
2x''+98x = -196sin(7t)
28 A2cos(7t) -28A1sin(7t) 

Therefore: A1 = 7 and A2 = 0. Substituting these values into the equation xp = A1tsin(7t) + A2tcos(7t) gives:

xp = 7tcos(7t)

Now solve for C1 and C2: since:

x(t)= xh + xp = C1cos(7t) + C2sin(7t) + 7tcos(7t)
and x(0)=0, x'(0)=7, we obtain
x(t) = 7tcos(7t)

b) sketch x(t), 0<t<1:

2. A 1 lb object is attached to a spring suspended from the ceiling. The spring has spring constant .5 and has an external force of sin(rt) acting on it. Find the r for which the displacement is in a state of pure resonance. (You do not have to find x(t).)

For pure resonance: r=sqrt(k/m)
We have k=.5  and m=1/32 .

r=sqrt(.5/ (1/32)) = sqrt(16) = 4

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, 

x(t)=C1cos(sqrt(k/m)t) +C2sin(sqrt(k/m)t)
Now we solve for the constants. We have x(0)= -2 = C1cos(0) + C2sin(0), therefore, C1= -2 . We have x'(0) = 1 , and since
x'(t) = -C1sin(t) +C2cos(t) , we have , x'(0) = 1 = 10C1sin(0) +10C2cos(0) , therefore, C2 = 1/10 , since k/m = 100:.
x(t) = -2cos(sqrt(10t) + (.1)sin(10t)

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

x(t) = sqrt(4.01)sin(10t -1.520837931) =(2.002498439)sin(10t -1.520837931).

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

t = .3091634258.

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.

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