Question 3.8.6: Interpret and solve the initial-value problem 1/5 d²x/dt² + ......

We can interpret the problem to represent a vibrational system consisting of a mass ( m = \frac{1}{5} slug or kilogram) attached to a spring (k = 2 \mathrm{lb} / \mathrm{ft} or \mathrm{N} / \mathrm{m}). The mass is released from rest \frac{1}{2} unit (foot or meter) below the equilibrium position. The motion is damped ( \beta 1.2) and is being driven by an external periodic (T = \pi / 2 \mathrm{~s}) force beginning at t = 0. Intuitively we would expect that even with damping, the system would remain in motion until such time as the forcing function was “turned off,” in which case the amplitudes would diminish. However, as the problem is given, f(t) = 5 \cos 4 t will remain “on” forever.

We first multiply the differential equation in (26) by 5 and solve

\frac{d x^{2}}{d t^{2}}+6 \frac{d x}{d t}+10 x = 0

by the usual methods. Since m_{1} =-3+i, m_{2} =-3-i, it follows that

x_{c}(t) = e^{-3 t}\left(c_{1} \cos t+c_{2} \sin t\right)

Using the method of undetermined coefficients, we assume a particular solution of the form x_{p}(t) = A \cos 4 t+B \sin 4 t. Differentiating x_{p}(t) and substituting into the DE gives

x_{p}^{\prime \prime}+6 x_{p}^{\prime}+10 x_{p} =(-6 A+24 B) \cos 4 t+(-24 A-6 B) \sin 4 t = 25 \cos 4 t.

The resulting system of equations

-6 A+24 B = 25, \quad-24 A-6 B = 0

yields A =-\frac{25}{102} and B = \frac{50}{51} . It follows that

x(t) = e^{-3 t}\left(c_{1} \cos t+c_{2} \sin t\right)-\frac{25}{102} \cos 4 t+\frac{50}{51} \sin 4 t.  (27)

When we set t = 0 in the above equation, we obtain c_{1} = \frac{38}{51}. By differentiating the expression and then setting t = 0, we also find that c_{2} =-\frac{86}{51}. Therefore the equation of motion is

x(t) = e^{-3 t}\left(\frac{38}{51} \cos t-\frac{86}{51} \sin t\right)-\frac{25}{102} \cos 4 t+\frac{50}{51} \sin 4 t    (28)