If d^(2)x/dt^(2)+g/b(x-a)=0,(a,b,g being positive constants) and x=a' and dx/dt=0 when t=0,show that x=a+(a'-a)cos{√(g)/(b)t}?

2 Answers
Jul 3, 2018

Answer:

# x(t) = (a'-a)cos(sqrt(g/b)t) + a \ \ \ \ #

Explanation:

We have:

# (d^2x)/(dt^2)+g/b(x-a)=0# with #x=a', dx/dt=0# when #t=0#

We can write the equation as:

# (d^2x)/(dt^2) + g/bx - (ag)/b = 0 iff (d^2x)/(dt^2) + g/bx = (ag)/b ..... [A]#

This is a second order non-Homogeneous Differentiation Equation. The standard approach is to find a solution, #y_c# of the homogeneous equation by looking at the Auxiliary Equation, which is the quadratic equation with the coefficients of the derivatives, and then finding an independent particular solution, #y_p# of the non-homogeneous equation.

Complementary Function

The homogeneous equation associated with [A] is

# x'' - 0x' +g/b = 0 #

And it's associated Auxiliary equation is:

# m^2 +g/b = 0 => m^2 = -g/b#

And as we are given that #a,b,g gt 0# then we have pure imaginary roots:

#m = +-sqrt(g/b)#

The roots of the auxiliary equation determine parts of the solution, which if linearly independent then the superposition of the solutions form the full general solution.

  • Real distinct roots #m=alpha,beta, ...# will yield linearly independent solutions of the form #x_1=Ae^(alphat)#, #x_2=Be^(betat)#, ...
  • Real repeated roots #m=alpha#, will yield a solution of the form #x=(At+B)e^(alphat)# where the polynomial has the same degree as the repeat.
  • Complex roots (which must occur as conjugate pairs) #m=p+-qi# will yield a pairs linearly independent solutions of the form # x=e^(pt)(Acos(qx)+Bsin(qx))#

Thus the solution of the homogeneous equation [A] is:

# x = e^(0x)(Acos(sqrt(g/b)t + Bsin(sqrt(g/b)t)) #
# \ \ = Acos(sqrt(g/b)t + Bsin(sqrt(g/b)t) #

Particular Solution

In order to find a particular solution of the non-homogeneous equation:

# (d^2x)/(dt^2) + g/bx = f(t) \ \ # with #f(t) = (ag)/b #

So, we should probably look for a solution of the form:

# x = C # ..... [B]

Where the constant #C# is to be determined by direct substitution and comparison:

Differentiating [B] wrt #x# twice we get:

# x' \ \= 0 #
# x'' = 0 #

Substituting these results into the DE [A] we get:

# (0) + g/bC = (ag)/b => C=a#

And so we form the Particular solution:

# x_p = a #

General Solution

Which then leads to the GS of [A}

# x(t) = x_c + x_p #
# \ \ \ \ \ \ \ = Acos(sqrt(g/b)t + Bsin(sqrt(g/b)t) + a#

Next we apply the initial conditions:

# x=a', dx/dt=0# when #t=0#

So that:

# a' = Acos0 + Bsin0 + a => A = a'-a#

And differentiating the above result wrt #t#

# x'(t) = -Asqrt(g/b)sin(sqrt(g/b)t) + Bsqrt(g/b)cos(sqrt(g/b)t) #

And again applying the initial conditions:

# 0 = -Asqrt(g/b)sin0 + Bsqrt(g/b)cos0 => B = 0 #

And so the complete solution is:

# x(t) = (a'-a)cos(sqrt(g/b)t) + a \ \ \ \ # QED

Jul 3, 2018

The Problem:

  • #x'' +g/b(x-a)=0 qquad qquad a,b,g gt 0#

  • #x_o=a' qquad x'_o =0#

Show that: #x=a+(a'-a)cos sqrt((g)/(b))t #

Substitute :

  • #bb(xi = x - a) qquad xi' = x' qquad xi'' = x''#

  • #xi_o=a' - a qquad xi'_o =0#

This makes it homogeneous , which means we can race to a solution.

  • #xi'' + omega^2 xi=0 qquad qquad omega^2 = g/b#

This is the Harmonic Oscillator, which can be solved in any number of ways, but always with well-known solution:

  • # {(xi =A cos omegat + B sin omega t), (xi' = - omega A sin omegat + omega B cos omega t):}#

Applying the IV's at #t = 0#:

  • # {(a' - a =A ), (0 = omega B ):}#

So:

#xi = (a' - a) cos sqrt(g/b)t#

Therefore:

#x = a + (a' - a) cos sqrt(g/b)t#