A 4-foot spring measures 8 feet long after a mass weighing 8 pounds is attached to it. The medium through which the mass moves offers a damping force numerically equal to sqrt(2) times the instantaneous velocity. A) Find the equation of motion if the mass

A) The equation of motion for the mass can be expressed as: {eq}m ddot{x} + c dot{x} + kx = mg {/eq} where m is the mass of the attached weight, x is the displacement of the weight from equilibrium position, c is the damping coefficient, k is the spring constant, and g is the acceleration due to gravity. Using the given values, we can substitute: {eq}(8)ddot{x} + (sqrt{2} \cdot dot{x}) + (k)(x) = (8)(32) {/eq} We can also find the spring constant k using the formula: {eq}k = \frac{mg}{x} {/eq} where m, g, and x are known values: {eq}k = \frac{(8)(32)}{4} = 64 {/eq} Substituting this into the equation of motion, we get: {eq}8ddot{x} + (sqrt{2} \cdot dot{x}) + (64)(x) = 256 {/eq} This is a second-order linear differential equation, which can be solved using standard methods such as the characteristic equation or Laplace transforms. B) To find the time at which the mass attains its extreme displacement, we need to find the point where the displacement is maximum. This occurs when the velocity is zero (at the turning point) and the acceleration is maximum (at the extremum). Setting {eq}dot{x} = 0 {/eq}, we get: {eq}8ddot{x} + (64)(x) = 256 {/eq} which simplifies to: {eq}ddot{x} + 8x = 32 {/eq} This is a homogeneous linear differential equation with constant coefficients, which can be solved using the auxiliary equation: {eq}r^2 + 8 = 0 {/eq} This gives the roots: {eq}r = \pm \sqrt{8}i {/eq} The general solution is therefore: {eq}x(t) = c_1 cos(sqrt{8}t) + c_2 sin(sqrt{8}t) {/eq} Using the initial condition {eq}x(0) = 0 {/eq}, we get: {eq}c_1 = 0 {/eq} Using the other initial condition {eq}dot{x}(0) = -7 {/eq}, we get: {eq}c_2 = -frac{7}{sqrt{8}} {/eq} Substituting these constants into the general solution, we get: {eq}x(t) = -frac{7}{sqrt{8}} sin(sqrt{8}t) {/eq} To find the time at which the mass attains its extreme displacement, we need to find the point where the sine function attains its maximum or minimum value. This occurs when: {eq}sqrt{8}t = \frac{pi}{2} + npi {/eq} where n is an integer (even for the maximum and odd for the minimum). Solving for t, we get: {eq}t_n = \frac{1}{sqrt{8}}left\frac{pi}{2} + npiright) {/eq} Substituting n = 3 (for the third turning point), we get: {eq}t_3 \approx 0.701 {/eq} C) To find the position of the mass at this instant, we simply substitute the value of t into the expression for x(t): {eq}x(t_3) = -frac{7}{sqrt{8}} sinleft(sqrt{8}\cdot 0.701right) \approx -1.566 {/eq} Therefore, the mass is at a displacement of about 1.566 feet below the equilibrium position at this instant.