Question

Show that if n>r>0, the complementary function represents vibrations which are ultimately damped out. Further, prove that the particular integral is of the form bcos(pt-α), where b^2= a^2/[(n^2-p^2)^2 +4r^2p^2].

Answer 1

$\displaystyle \frac{\mathrm{d}^2x}{\mathrm{d}t^2} + 2r\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)+ n^2x= a\cos(pt) \\ \textsf{Let}\, x = f(t) \\ f"(t) + 2rf'(t) + n^2f(t) = a\cos(pt)\\ \textsf{The complementary function is derived}\\ \textsf{when}\, f"(t) + 2rf'(t) + n^2f(t) = 0\\ \text{The auxiliary equation is}\\ m^2 + 2rm + n^2 = 0\\ m = \frac{-2r \pm \sqrt{4r^2 - 4n^2}}{2} = -r \pm \sqrt{r^2 - n^2}\\ \textsf{Since}\, n > r\\ m = -r \pm j\sqrt{n^2 - r^2}\\ \textsf{Recall that if the auxiliary equation is}\\ m = \alpha \pm j\beta, \, \textsf{the general solution can be}\\ \textsf{written as}\\ y = e^{\alpha t}(A\cos(\beta t) + B\sin(\beta t))\\ \therefore \textsf{The complementary function of the given DE is}\\ f(t) = e^{-rt}(C\cos(\sqrt{n^2 - r^2}t + D\sin(\sqrt{n^2 - r^2}))\\ \textsf{This is a system with a frequency of}\\ \frac{2\pi}{\sqrt{n^2 - r^2}}\, \textsf{per units of time and a}\\ \textsf{period of}\, \sqrt{n^2 - r^2} \, \textsf{units of time and an amplitude}\\ \textsf{that is decreasing with time.}\\ \textsf{The effect of the}\, f'(t)\, \textsf{in the differential}\\ \textsf{equation is to introduce damping}\\\textsf{into the oscillatory motion so}\\\textsf{causing it to decay.}\\ \textsf{The coefficient}\, e^{-rt} \, \textsf{introduced damping}\\ \textsf{into the system, and since}\\ \textsf{as}\, t \rightarrow \infty, e^{-rt} \rightarrow 0 \implies f(t)\rightarrow 0,\,\textsf{the vibrations}\\ \textsf{are ultimately damped out.}\\ \textsf{The particular integral is of the form}\\ f(t) = A\cos(pt) + B\sin(pt)\\ f'(t) = -Ap\sin(pt) + Bp\cos(pt)\\ f''(t) = -Ap^2\cos(pt) - Bp^2\sin(pt)\\ -Ap^2\cos(pt) - Bp^2\sin(pt) + 2r(-Ap\sin(pt) + Bp\cos(pt)) + n^2(A\cos(pt) + B\sin(pt)) = a\cos(pt)\\ (-Ap^2 + 2rBp + An^2)\cos(pt) + (-Bp^2 - 2rAp + Bn^2)\sin(pt) = a\cos(pt)\\ \textsf{Comparing coefficients}\\ -Ap^2 + 2rBp + An^2 = a\\ -Bp^2 - 2rAp + Bn^2 = 0\\ A(n^2 - p^2) + 2rBp = a\hspace{0.5cm}(1)\\ B(n^2 - p^2) - 2rAp = 0\hspace{0.5cm}(2)\\ (1) \times B \, \textsf{yields}\\ AB(n^2 - p^2) + 2rB^2p = aB\hspace{0.5cm}(3)\\ (2) \times A \, \textsf{yields}\\ AB(n^2 - p^2) - 2rA^2p = 0\hspace{0.5cm}(4)\\ \textsf{Subtract}\, (4) \, \textsf{from}\, (3)\\ 2rA^2 + 2rB^2 = aB\\ A^2(n^2 - p^2) + 2rABp = aA\hspace{0.5cm}(5)\\ B^2(n^2 - p^2) - 2rABp = 0\hspace{0.5cm}(6)\\ \textsf{Add}\, (5)\, \textsf{and}\, (6)\\ (A^2 + B^2)(n^2 - p^2) = aA\hspace{0.5cm}(7)\\ 2r(A^2 + B^2) = aB\hspace{0.5cm}(8)\\ \implies A^2 + B^2 = \frac{aB}{2r}\\ \textsf{Inserting in}\, (7)\\ \frac{aB}{2r}\cdot(n^2 - p^2) = aA\\ B(n^2 - p^2) = 2rA\\ \implies B = \frac{2rA}{n^2 - p^2}\\ \textsf{By}\, (5) \,\textsf{and} \, (6)\\ A + \frac{2rpB}{n^2 - p^2} = \frac{a}{n^2 - p^2}\\ -\frac{2rpA}{n^2 - p^2} + B = 0\\ -\frac{4rp^2 A}{(n^2 - p^2)^2} + \frac{2rpB}{n^2 - p^2} = 0\\ A\left(1 + \frac{(2rp)^2}{(n^2 - p^2)}\right) = \frac{a}{n^2 - p^2}\\ A\left(\frac{(n^2 - p^2)^2 + 4r^2p^2}{(n^2 - p^2)^2}\right) = a\\ A = \frac{a(n^2 - p^2)}{(n^2 - p^2)^2 + 4r^2p^2}\\ B =\frac{2r}{n^2 - p^2}\\ A = \frac{2arp}{(n^2 - p^2)^2 + 4r^2p^2}\\ \therefore f(t) = \frac{a(n^2 - p^2)}{(n^2 - p^2)^2 + 4r^2p^2}\cos(pt) + \frac{2arp}{(n^2 - p^2)^2 + 4r^2p^2}\sin(pt)\\ \begin{aligned} f(t) &= \frac{a}{(n^2 - p^2)^2 + 4r^2p^2}\sqrt{(2rp)^2 + (n^2 - p^2)}\cos\left(pt - \arctan\left(\frac{2r}{n^2 - p^2}\right)\right)\\ &=\frac{a}{\sqrt{(n^2 - p^2)^2 + 4r^2p^2}}\cos\left(pt - \arctan\left(\frac{2r}{n^2 - p^2}\right)\right) \end{aligned}\\ \textsf{Thus, the Particular integral is of the form}\\ b\cos(pt - \alpha),\, \textsf{where}\, b^2 = \frac{a^2}{(n^2 - p^2)^2 + 4r^2p^2}\\ \textsf{And the solution to the differential equation is}\\ x = e^{-rt}(C\cos(\sqrt{n^2 - r^2}t + D\sin(\sqrt{n^2 - r^2})) + \frac{a}{\sqrt{(n^2 - p^2)^2 + 4r^2p^2}}\cos\left(pt - \arctan\left(\frac{2r}{n^2 - p^2}\right)\right)\\$