A rope of length L is attached to the ceiling and make small vibrations. The dis

Question

A rope of length L is attached to the ceiling and make small vibrations. The displacement from vertical at any point x is given by u(t, x). a. Consider a point that is a distance x from the ceiling. Explain why the weight of the cable below this point is given by W = rho g(L - x) where rho is the linear density (mass per unit length) b. Given that the restoring force can be approximated by F(x) almostequalto Wu_x use the definition of the derivative of F(x) to explain why F(x + Delta x) - F(x) = Delta x(Wu_x)_x c. Next, use Newton's law and the above results to explain why rho Delta xu = Delta x rho g [(L - x) u_x]_x d. Finally assume u(t, x) y(x) cos(omega t + delta) and substitute this into the above formula to derive the ODE (L - x) y" - y' + omega^2/g y = 0 This ODE can then be written as Bessel's equation via a change of variables (see the Critical Thinking for this section, The resulting ODE is xy" + y' + xy = 0. Solve this using the series solution methods discussed in this experience. Write out the first 5 non-zero coefficients for the series below. Assume initial conditions y(0) = 1, y'(0) = 0

Explanation / Answer

a. consider a point at distance x from the ceiling
length of rope below this point = L - x
linear density of rope = rho
mass of length L = rho*L
mass of length (L - x) = rho*(L - x)
weight of this mass = rho*g(L - x)

b. now, assuming F(x) = W*du/dx
then from first principle
dF(x)/dx = [F(x + dx) - F(x)]/dx
but dF(x)/dx = d(W*du/dx)/dx
so, F(x + dx) - F(x) = dx[d(W*du/dx)/dx]

c. From nerwton's second law
F(x + dx) - F(x) = dm*Utt [ dm is mass of the segment of length dx at point x, and Utt is double time derivative of U at that point]
so, dm = rho*dx
hence
rho*dx*Utt = dx[d(W*du/dx)/dx] = dx[d(rho*g(L-x)*du/dx)/dx] = dx*rho*g[d((L-x)*du/dx)/dx]
d. let, u(t,x) = y(x)cos(wt + del)
Ut = -y(x)wsin(wt + del)
Utt = -y(x)*w^2*cos(wt + del)
Ux = y'(x)cos(wt + dwl)

so, substituting
Utt = g[d((L-x)*du/dx)/dx]
-y(x)*w^2*cos(wt + del) = g[d((L-x)*y'(x)cos(wt + dwl))/dx]
-y(x)*w^2 = g[d((L-x)*y'(x))/dx] = g[Ly"(x) - xy"(x) - y'(x)]
(L - x)y" - y' + w^2*y/g = 0

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