# circular motion and the equations of SHM

To explore the properties of simple harmonic motion, we must express the displacement x of the oscillating body as a function of time, x(t). The second derivative of this function, d2 x/dt2 , must be equal to 1-k/m2 times the function itself, as required by Eq. (14.4). As we mentioned, the formulas for constant acceleration from Section 2.4 are no help because the acceleration changes constantly as the displacement x changes. Instead, we’ll find x(t) by noting that SHM is related to uniform circular motion, which we studied in Section 3.4

Figure 14.5a shows a top view of a horizontal disk of radius A with a ball attached to its rim at point Q. The disk rotates with constant angular speed v (measured in rad/s), so the ball moves in uniform circular motion. A horizontal light beam casts a shadow of the ball on a screen. The shadow at point P oscillates back and forth as the ball moves in a circle. We then arrange a body attached to an ideal spring, like the combination shown in Figs. 14.1 and 14.2, so that the body oscillates parallel to the shadow. We will prove that the motions of the body and of the ball’s shadow are identical if the amplitude of the body’s oscillation is equal to the disk radius A, and if the angular frequency 2pf of the oscillating body is equal to the angular speed v of the rotating disk. That is, simple harmonic motion is the projection of uniform circular motion onto a diameter.

We can verify this remarkable statement by finding the acceleration of the shadow at P and comparing it to the acceleration of a body undergoing SHM, given by Eq. (14.4). The circle in which the ball moves so that its projection matches the motion of the oscillating body is called the reference circle; we will call the point Q the reference point. We take the reference circle to lie in the xy-plane, with the origin O at the center of the circle (Fig. 14.5b). At time t the vector OQ from the origin to reference point Q makes an angle u with the positive x-axis. As point Q moves around the reference circle with constant angular speed v, vector OQ rotates with the same angular speed. Such a rotating vector is called a phasor. (This term was in use long before the invention of the Star Trek stun gun with a similar name.) We’ll use phasors again when we study alternating-current circuits in Chapter 31 and the interference of light in Chapters 35 and 36.

The x-component of the phasor at time t is just the x-coordinate of the point Q:

x = Acos Ø                       (14.5)

This is also the x-coordinate of the shadow P, which is the projection of Q onto the x-axis. Hence the x-velocity of the shadow P along the x-axis is equal to the x-component of the velocity vector of point Q (Fig. 14.6a), and the x-acceleration of P is equal to the x-component of the acceleration vector of Q (Fig. 14.6b). Since point Q is in uniform circular motion, its acceleration vector a S Q is always directed toward O. Furthermore, the magnitude of a S Q is constant and given by the angular speed squared times the radius of the circle (see Section 9.3):

Figure 14.6b shows that the x-component of a ¯ Q is ax = -aQ cosØ. Combining this with Eqs. (14.5) and (14.6), we get that the acceleration of point P is

The acceleration of point P is directly proportional to the displacement x and always has the opposite sign. These are precisely the hallmarks of simple harmonic motion.

Equation (14.8) is exactly the same as Eq. (14.4) for the acceleration of a harmonic oscillator, provided that the angular speed v of the reference point Q is related to the force constant k and mass m of the oscillating body by

We have been using the same symbol v for the angular speed of the reference point Q and the angular frequency of the oscillating point P. The reason is that these quantities are equal! If point Q makes one complete revolution in time T, then point P goes through one complete cycle of oscillation in the same time; hence T is the period of the oscillation. During time T the point Q moves through 2p radians, so its angular speed is v = 2p/T. But this is the same as Eq. (14.2) for the angular frequency of the point P, which verifies our statement about the two interpretations of v. This is why we introduced angular frequency in Section 14.1; this quantity makes the connection between oscillation and circular motion. So we reinterpret Eq. (14.9) as an expression for the angular frequency of simple harmonic motion:

When you start a body oscillating in SHM, the value of v is not yours to choose; it is predetermined by the values of k and m. The units of k are N/m or kg/s 2 , so k>m is in (kg>s 2 )/kg = s -2 . When we take the square root in Eq. (14.10), we get s -1 , or more properly rad/s because this is an angular frequency (recall that a radian is not a true unit).

According to Eqs. (14.1) and (14.2), the frequency f and period T are

We see from Eq. (14.12) that a larger mass m will have less acceleration and take a longer time for a complete cycle (Fig. 14.7). A stiffer spring (one with a larger force constant k) exerts a greater force at a given deformation x, causing greater acceleration and a shorter time T per cycle.

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Ans: The simplest kind of oscillation occurs when the restoring force Fx is directly proportional to the displacement from equilibrium x. This happens if the spring in Figs. 14.1 and 14.2 is an ideal one that obeys Hooke’s law view more..
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Ans: Here are some terms that we’ll use in discussing periodic motions of all kinds: view more..
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Ans: n. A body with mass m rests on a frictionless horizontal guide system, such as a linear air track, so it can move along the x-axis only. The body is attached to a spring of negligible mass that can be either stretched or compressed. The left end of the spring is held fixed, and the right end is attached to the body. The spring force is the only horizontal force acting on the body; the vertical normal and gravitational forces always add to zero view more..
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Ans: To explore the properties of simple harmonic motion, we must express the displacement x of the oscillating body as a function of time, x1t2. view more..
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Ans: the period and frequency of simple harmonic motion are completely determined by the mass m and the force constant k. In simple harmonic motion the period and frequency do not depend on the amplitude A. view more..
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Ans: We still need to find the displacement x as a function of time for a harmonic oscillator. Equation (14.4) for a body in SHM along the x-axis is identical to Eq. (14.8) for the x-coordinate of the reference point in uniform circular motion with constant angular speed v = 2k/m view more..
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Ans: We can learn even more about simple harmonic motion by using energy considerations. The only horizontal force on the body in SHM in Figs. 14.2 and 14.13 is the conservative force exerted by an ideal spring. The vertical forces do no work, so the total mechanical energy of the system is conserved. We also assume that the mass of the spring itself is negligible. view more..
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Ans: the energy quantities E, K, and U at x = 0, x = ±A/2, and x = ±A. Figure 14.15 is a graphical display of Eq. (14.21); energy (kinetic, potential, and total) is plotted vertically and the coordinate x is plotted horizontally. The parabolic curve in Fig. 14.15a represents the potential energy U = 1/2 kx2 . The horizontal line represents the total mechanical energy E, which is constant and does not vary with x. view more..
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Ans: PROBLEM SOLVING STRATEGY ON ENERGY MOMENTUM OF SHM view more..
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Ans: So far, we’ve looked at a grand total of one situation in which simple harmonic motion (SHM) occurs: a body attached to an ideal horizontal spring. But SHM can occur in any system in which there is a restoring force that is directly proportional to the displacement from equilibrium, as given by Eq. (14.3), Fx = -kx view more..
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Ans: A mechanical watch keeps time based on the oscillations of a balance wheel (Fig. 14.19). The wheel has a moment of inertia I about its axis. A coil spring exerts a restoring torque tz that is proportional to the angular displacement u from the equilibrium position. We write tz = -ku, where k (the Greek letter kappa) is a constant called the torsion constant. Using the rotational analog of Newton’s second law for a rigid body, gtz = Iaz = I d2 u>dt2 view more..
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Ans: The following discussion of the vibrations of molecules uses the binomial theorem. If you aren’t familiar with this theorem, you should read about it in the appropriate section of a math textbook. view more..
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Ans: A simple pendulum is an idealized model consisting of a point mass suspended by a massless, unstretchable string. When the point mass is pulled to one side of its straight-down equilibrium position and released, it oscillates about the equilibrium position. view more..
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Ans: A physical pendulum is any real pendulum that uses an extended body, as contrasted to the idealized simple pendulum with all of its mass concentrated at a point. F view more..
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