Consider the system of two masses and three springs shown in the figure with the
ID: 2967130 • Letter: C
Question
Consider the system of two masses and three springs shown in the figure
with the following equations of motion:
and the following masses and spring constants
Find the general solution of the above system using the given masses and spring constants by solving the system and applying the initial conditions x(0)=A, x'(0)=B
y(0)=C and y'(0)=D
Consider the system of two masses and three springs shown in the figure Find the general solution of the above system using the given masses and spring constants by solving the system and applying the initial conditions x(0)=A, x'(0)=B y(0)=C and y'(0)=D and the following masses and spring constants with the following equations of motion:Explanation / Answer
m1*x'' = ?(k1+k2)*x + k2*y = -k1*x - k2*(x - y)
m2*y'' = k2*x ? (k2+k3)*y = -k3*y - k2*(y - x)
[Note that the left hand side involves the second derivatives of x and y; you left out the "prime" symbols indicating this in your question.]
Plugging in the values for the parameters, we get:
x'' = -x - 2(x-y)
y'' = -y + 2(x-y)
Adding these equations together yields:
x'' + y'' = (x+y)'' = -(x+y)
and subtracting the the equation for y'' from the equation for x'' yields:
x'' - y'' = (x-y)'' = -5(x-y)
Define the new variables (called normal coordinates) u = (x+y) and v = (x - y) and we get the decoupled equations:
u'' + u = 0
and
v'' + 5v = 0
You should recognize these as the equations of motion for simple, undamped, unforced harmonic oscillators. The equation for u has angular frequency 1, and the equation for v has angular frequency sqrt(5). These are the frequencies of the "normal modes" of oscillation of this coupled system of oscillators.
The solutions for u and v are given by:
u(t) = a*cos(t) + b*sin(t) and v(t) = c*cos(sqrt(5)*t) + d*sin(sqrt(5)*t)
where a, b, c, and d constants of integration.
Back substituting for u and v, we have:
(x + y) = a*cos(t) + b*sin(t)
(x - y) = c*cos(sqrt(5)*t) + d*sin(sqrt(5)*t)
Alternately adding and subtracting these equations from one another, we get:
x = (1/2)*[a*cos(t) + b*sin(t) + c*cos(sqrt(5)*t) + d*sin(sqrt(5)*t)]
and
y = (1/2)*[a*cos(t) + b*sin(t) - c*cos(sqrt(5)*t) - d*sin(sqrt(5)*t)]
I assume you meant to write that the initial conditions are: x(0)=A, x'(0)=B, y(0)=C and y'(0)=D [note primes indicating the first derivatives].
From x(0) = A and y(0) = c, we get:
x(0) = A = (1/2)*[a + c]
and
y(0) = C = (1/2)*[a - c]
so
a = A + C and c = A - C
Taking the derivatives of our solutions:
x' = (1/2)*[-a*sin(t) + b*cos(t) - c*sqrt(5)*sin(sqrt(5)*t) + d*sqrt(5)*cos(sqrt(5)*t)]
y' = (1/2)*[-a*sin(t) + b*cos(t) + c*sqrt(5)*sin(sqrt(5)*t) - d*sqrt(5)*cos(sqrt(5)*t)]
x'(0) = B = (1/2)*( b + d*sqrt(5))
and
y'(0) = D = (1/2)*(b - d*sqrt(5))
b = B+D and d = (B-D)/sqrt(5)
so the solution to the IVP is:
x = (1/2)*[(A+C)*cos(t) + (B+D)*sin(t) + (A - C)*cos(sqrt(5)*t) + ((B-D)/sqrt(5))*sin(sqrt(5)*t)]
and
y = (1/2)*[(A+C)*cos(t) + (B+D)*sin(t) - (A - C)*cos(sqrt(5)*t) - ((B-D)/sqrt(5))*sin(sqrt(5)*t)]