In this last chapter of the course, we handled two physical phenomena which involve a linear second order, constant coefficients, differential equations, the spring mass system, and the motion of the pendulum. First, let's consider the spring mass system. Attach a mass m to a spring of length l, which is suspended from a rigid support so that the spring is stretched with elongation Delta of l, and reaches its equilibrium state. Then there are two forces acting on the mass. Say the gravitational force, that is, the weight w is equal to m times g, m is the mass of the object and g is the gravitational acceleration, which acting downward, and the spring's restoring force F_s acting upward. Here we adopt the convention that the downward direction is positive. Here's the unstretched spring, then we put the mass m, then we have an elongation given by the Delta of l. Then by the Hooke's law, when Delta l is small compared to the original length l, the restoring force F_s is proportional to the elongation. In other words, F_s is equal to negative k times Delta l, where the positive constant k is the so-called spring constant or the stiffness of the spring. Since the mass is at its equilibrium state, we should have mg minus k times Delta l is equal to 0. In other words, mg is equal to the gravitational force, is the same as the restoring force k times Delta l. Now let the mass be displaced further from its equilibrium state as shown in this figure by an amount x of t, which is measured positive downward and then released. We are going to look at the motion of this mass. Then actually, there are four forces acting on the system. First, the weight w is equal to m times g due to the gravitation and which is acting downward, and the spring force F_s, which is equal to minus k times Delta plus x, which is acting upward when the spring is extended. In other words, the Delta plus x is positive or the spring force is acting downward when the spring is compressed. In other words, Delta l plus x is negative. Another force that we have is the damping force F_d due to the viscosity of the fluid in which the mass is moving, which is proportional to the mass' of speed so that we have F_d is equal to negative c times x of prime, where the positive constant c is the damping constant and the x prime is the velocity of the object, when the speed of the mass is quite small. Here, the negative sign means that the damping force is acting in the direction opposite to the mass' moving direction. Finally, there might be some external force f of t acting on the mass. Then by the Newton's second law of motion, first that we can think of the forces of the system is m times acceleration. So m times x double prime must be equal to the total force acting on the system. In other words, if we have the gravitational force mg and the spring force F_s and the damping force F_d and any external force f, that is equal to m times g minus k times Delta l plus x minus c times x prime plus f. But since we know that at the equilibrium state mg is equal to k times Delta l, so mg minus k Delta l canceled out. Then simplifying the equation, we get an x double prime plus cx prime plus kx is equal to f, which is a second-order linear constant coefficients, possibly non-homogeneous differential equation. We call this differential equation 1, as the differential equation for forced damped motion for general f. If f is identically zero, then we call it as free damped motion. Divide the whole equation by m, then you are going to get x double prime plus c over m x prime plus k over m and x, and that is equal to f over m. You get this equation. Next set for the computation or simplicity, let's recall c over m is equal to 2 times Lambda and the k over m let's say it equal to Omega squared and f of n let's say it to be g, then we get the second equation. Say x double prime plus 2 Lambda times x prime plus Omega squared x is equal to g of t.
