21-3 Harmonic motion and circular motion The fact that cosines are involved in the solution of Eq.( 21.2) there might be some relationship to circles. This is artificial, of course, because there is no circle actually involved in the linear motion-it just goes up and down We may point out that we have, in fact, already solved that equation when we were studying the mechanics of circular motion. If a particle moves in a circle with a constant speed v, the radius vector from the center of the circle to the particle turns through an angle whose size is proportional to the time. If we call this angle 0= vt/R(Fig. 21-2)then de/dt v/R. We know that there an acceleration a =u2/R=oR toward the center. Now we also know that the position x, at a given moment, is the radius of the circle times cos 0, and that y is the radius times sin 0 A particle moving in a circular path at constant speed x = r cos e, y= R sin 8. Now what about the acceleration? What is the x-component of acceleration d2x/dt2? We have already worked that out geometrically; it is the magnitude of the acceleration times the cosine of the projection angle, with a minus sign because it is toward the center (21.7) In other words, when a particle is moving in a circle, the horizontal component of its motion has an acceleration which is proportional to the horizontal displacement from the center. of course we also have the solution for motion in a circle x= R cos Wot. Equation (21.7)does not depend upon the radius of the circle so for a circle of any radius, one finds the same equation for a given wo. Thus, r several reasons, we expect that the displacement of a mass on a spring will turn out to be proportional to cos wof, and will, in fact, be exactly the same motion as we would see if we looked at the x-component of the position of an object rotating in a circle with angular velocity wo. As a check on this, one can devise an experi ment to show that the up-and-down motion of a mass on a spring is the same that of a point going around in a circle. In Fig. 21-3 an arc light projected on a screen casts shadows of a crank pin on a shaft and of a vertically oscillating ma side by side. If we let go of the at the right ig.21-3. Demonstration of the the shaft speed is carefully adjusted so that the frequencies match, each should equivalence between simple harmonic follow the other exactly. One can also check the numerical solution we obtained motion and uniform circular motion earlier with the cosine function, and see whether that agrees very well Here we may point out that because uniform motion in a circle is so closely related mathematically illatory up-and-down motion, we can analyze oscillatory motion in a simpler way if we imagine it to be a projection of something going in a circle. In other words, although the distance y means nothing in the oscillator problem, we may still artificially supplement Eq.(21. 2)with another equation using y, and put the two together. If we do this, we will be able to analyze our one-dimensional oscillator with circular motions, which is a lot easier than having to solve a differential equation. The trick in doing this is to use complex numbers, a procedure we shall introduce in the next chapter. 21-4 Initial conditions Now let us consider what determines the constants A and B or a and 4. of course these are determined by how we start the motion If we start the motion with just a small displacement, that is one type of oscillation; if we start with an initial displacement and then push up when we let go still a different motion. The constants A and B, or a and A, or any other way of putting it, are determined, of course, by the way the motion started, not by any other features of the situation. These are called the initial conditions. We would like to connect the initial conditions with the constants. Although this can be done using any one of the forms(21.6), it turns out to be easiest if we use Eq.( 21.6c). Suppose that at t=0 we have started with an initial displacement xo and a certain velocity vo