First consider the most simple case, say, the free undamped motion. By which I mean, there is no external force. In other words, f is identical to 0. And there is no damping, so that the damping constant c is equal to 0, okay? Then the given differential equation becomes, simpler one, x double prime + omega square times x = 0. And this general solution is very easy to find because the characteristic equation corresponding to this differential equation is r2 + omega squared = 0, so that we have two roots for r given by +or- i times omega. So that its general solution becomes x(t) = arbitrary constant c sub 1 times cosine omega t, and + another arbitrary constant c sub 2 times sine omega t, all right? But using the trigonometrical identity, we can combine these two terms into one as capital A times cosine (omega t- phi), where A = square root of c1 squared + c2 squared. And the c1 = capital A times the cosine phi, c2 = A times the sine phi, okay? Here, we call the capital A, given by this quantity, the amplitude. And the angle phi, the phase angle, of the simple harmonic motion (4). We call this the equation. The equation 4 has a simple harmonic motion and where we call the A, the amplitude and phi the phrase angle, okay? Note that there is a simple harmonic motion given by the equation 4, has period T is equal to 2 pi omega. And if we can see 1 over t, that is equal to, F, is equal to omega over 2 pi, right? Let's look at the picture for the simple harmonic motion, okay? It's a kind of cosine curve,okay? Where we have the maximum altitude, capital A, this is amplitude, right? And where the maximum values obtained when t is equal to phi over omega, okay? And from this, the length of the consecutive two maximum points, that is equal to period 2 pi over omega, right? That's the graph of the simple harmonic motion, okay? For example, let's assume that we have an 1kg mass is attached to a spring with stiffness 4 kg per second square. And at time t = 0, the mass is stretched downward scaled to 3 over 4 meters from its equilibrium point and then released with upward velocity one-half meters per second, okay? Assuming there is no damping and no external force, find its motion and the amplitude, and the phase angle in -pi and pi, okay? Also find the earliest time after release, at which the mass passes through the equilibrium position, okay? And the problem says, the motion is a free undamped motion because there is no external forces free and because we assumed that there is no damping, so it's undamped too, right? And m = 1, and the stiffness, k = 4, right? So that we have an initial value problem, very simple one, x double prime + 4x = 0. Initial displacement is the square root of 3 over 4. And the initial velocity, because it is moving outward, it should be- one-half, right? Solving this very simple homogeneous second order, constant coefficient differential equation. With these two initial conditions, you can easily get x(t) = square root of 3 over 4 cosine 2t- 1 over 4 sine 2t, okay? And using the trigonometric identity, we can combine it into a one-half times cosine 2t + pi over 6, right? One-half amplitude you get the following. This is equal to c1 squared, in other words, the square root of 3 over 4 squared and + (- 1 over 4)2, right? That's the amplitude one-half, okay? And the angle is equal to, you get A is equal to one-half times the cosine of phi, that is equal square root of 3 over 4, and one-half of sine of phi, that is equal to -1 over 4. From this, you can get the phase angle, phi is equal to- pi over 6. The mass passes through the, that's the harmonium motion we obtain, okay? And finally, to get the moment at which the mass passes through the equilibrium position, and equilibrium position means that the displacement x(t) = 0. So let's put x(t), say, one-half times cosine 2t + pi over 6 = 0, then cosine is equal to 0 when the argument is pi over 2 + k pi, where k is any integer. So that we should have 2t + pi over 6 must be equal to pi over 2 plus k pi, where k is an arbitrary integer. So first such time, first such positive time is when k is equal to 0, from this expression, you will get pi over 6, right? So that means after pi over 6 seconds, the mass passes through the equilibrium position, in other words, at which x = 0, right?
