Frictional oscillations -2

The forces on the block in the $x$ -direction are a component of its weight $Mg \sin \theta$ , the spring force $-kx$ , and friction $\pm mg \mu \cos \theta$ . Notice that the net work done on the block from $x_n$ to $x_{n + 1}$ is 0 because there is no net change in the block's velocity, so $\int_{x_n}^{x_{n + 1}} (Mg \sin \theta \pm Mg \mu \cos \theta - kx)\,dx = 0.$ When we solve this integral equation for $x_{n + 1}$ , we obtain the recurrence relation $x_{n + 1} = -x_n + \frac {2Mg}k(\sin \theta \pm \mu \cos \theta).$ The sign of $\mu \cos \theta$ alternates. When $n = 0$ , friction and gravity are working against each other so $\mu \cos \theta$ is negative. In general, we can write $x_{n + 1} = -x_n + \frac {2Mg}k \big[ \sin \theta + (-1)^{n + 1} \mu \cos \theta \big].$ In terms of $a$ , $b$ , $c$ , and $d$ , this recurrence relation is $\frac {Mg}k(-1)^{n + 1}\big[a(b^{n + 1} - c)\sin \theta + d\mu(n + 1)\cos \theta \big] = -\frac {Mg}k(-1)^n \big[ a(b^n - c)\sin \theta + d\mu n \cos \theta \big] + \frac {2Mg}k \big[\sin \theta + (-1)^{n + 1} \mu \cos \theta \big].$ When we gather the $\sin \theta$ terms on the right and equate the coefficients of $\sin \theta$ on the left and right, we obtain $\frac {Mg}k (-1)^{n + 1} a(b^{n + 1} - c) = \frac {Mg}k(-1)^n \big[ 2(-1)^n - a(b^n - c) \big],$ which simplifies to $-ab^n(b - 1) = 2(-1)^n.$ This equation holds for all $n$ only if $b = -1 \implies -a(-1)^n(-1 - 1) = 2(-1)^n \implies a = 1.$ When we equate the coefficients of $\cos \theta$ from the two sides of the recurrence, we obtain $\frac {Mg}k(-1)^{n + 1} d\mu(n + 1) = \frac {Mg}k(-1)^{n + 1}(2\mu + d\mu n),$ which simplifies to $d(n + 1) = 2 + dn \implies d = 2.$ Now we have found that $a = 1$ , $b = -1$ , and $d = 2$ . To find $c$ , recall that the block is released from rest when the spring is at its natural length, so $0 = x_0 = \frac {Mg}k(-1)^0 \Big( 1 \big[ (-1)^0 - c \big] \sin \theta + 2\mu(0) \cos \theta \Big) = \frac{Mg}k(1 - c)\sin \theta.$ The first and last parts of this equation imply $\frac {Mg}k(1 - c)\sin \theta = 0 \implies c = 1.$ Therefore, we have found $a = c = 1$ , $b = -1$ , $d = 2$ , and $a^2 + b^2 + c^2 + d^2 + a + b + c + d = \boxed{10}.$