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15-3 the Force Law for Simple Harmonic Motion

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15-3 the Force Law for Simple Harmonic Motion 390 CHAPTER 15 OSCILLATIONS Thus, as Fig. 15-4 shows, when the displacement has its greatest positive value, the acceleration has its greatest negative value, and conversely. When the displace- ment is zero, the acceleration is also zero. 15-3 The Force Law for Simple Harmonic Motion Once we know how the acceleration of a particle varies with time, we can use Newton’s second law to learn what force must act on the particle to give it that acceleration. If we combine Newton’s second law and Eq. 15-8, we find, for sim- ple harmonic motion, F ϭ ma ϭϪ(m␻2)x. (15-9) This result—a restoring force that is proportional to the displacement but oppo- site in sign—is familiar. It is Hooke’s law, F ϭϪkx, (15-10) for a spring, the spring constant here being k ϭ m␻2. (15-11) We can in fact take Eq. 15-10 as an alternative definition of simple harmonic motion. It says: Simple harmonic motion is the motion executed by a particle subject to a force that is proportional to the displacement of the particle but opposite in sign. The block–spring system of Fig. 15-5 forms a linear simple harmonic k oscillator (linear oscillator, for short), where “linear” indicates that F is pro- m portional to x rather than to some other power of x. The angular frequency ␻ of x the simple harmonic motion of the block is related to the spring constant k and –xm x = 0 +xm the mass m of the block by Eq. 15-11, which yields Fig. 15-5 A linear simple harmonic os- cillator.The surface is frictionless. Like the k ␻ ϭ (angular frequency). (15-12) particle of Fig. 15-1, the block moves in sim- A m ple harmonic motion once it has been ei- ther pulled or pushed away from the x ϭ 0 By combining Eqs. 15-5 and 15-12, we can write, for the period of the linear position and released. Its displacement is oscillator of Fig. 15-5, then given by Eq. 15-3. m T ϭ 2␲ (period). (15-13) A k Equations 15-12 and 15-13 tell us that a large angular frequency (and thus a small period) goes with a stiff spring (large k) and a light block (small m). Every oscillating system, be it a diving board or a violin string, has some element of “springiness” and some element of “inertia” or mass, and thus resem- bles a linear oscillator. In the linear oscillator of Fig. 15-5, these elements are located in separate parts of the system: The springiness is entirely in the spring, which we assume to be massless, and the inertia is entirely in the block, which we assume to be rigid. In a violin string, however, the two elements are both within the string, as you will see in Chapter 16. CHECKPOINT 2 Which of the following relationships between the force F on a particle and the particle’s position x implies simple harmonic oscillation: (a) F ϭϪ5x, (b) F ϭϪ400x2, (c) F ϭ 10x, (d) F ϭ 3x2? PART 2 15-3 THE FORCE LAW FOR SIMPLE HARMONIC MOTION 391 Sample Problem Block–spring SHM, amplitude, acceleration, phase constant A block whose mass m is 680 g is fastened to a spring whose Calculation: Thus, we have spring constant k is 65 N/m. The block is pulled a distance v ϭ ␻x ϭ (9.78 rad/s)(0.11 m) x ϭ 11 cm from its equilibrium position at x ϭ 0 on a fric- m m tionless surface and released from rest at t ϭ 0. ϭ 1.1 m/s. (Answer) (a) What are the angular frequency, the frequency, and the This maximum speed occurs when the oscillating block is rush- period of the resulting motion? ing through the origin; compare Figs. 15-4a and 15-4b, where you can see that the speed is a maximum whenever x ϭ 0. KEY IDEA (d) What is the magnitude am of the maximum acceleration of the block? The block–spring system forms a linear simple harmonic oscillator, with the block undergoing SHM. KEY IDEA Calculations: The angular frequency is given by Eq. 15-12: The magnitude am of the maximum acceleration is the accel- ␻2 eration amplitude xm in Eq. 15-7. k 65 N/m ␻ ϭ ϭ ϭ 9.78 rad/s A m A 0.68 kg Calculation: So, we have ϭ ␻2 ϭ 2 Ϸ 9.8 rad/s. (Answer) am xm (9.78 rad/s) (0.11 m) ϭ 2 The frequency follows from Eq. 15-5, which yields 11 m/s . (Answer) ␻ 9.78 rad/s This maximum acceleration occurs when the block is at the f ϭ ϭ ϭ 1.56 Hz Ϸ 1.6 Hz. (Answer) ends of its path. At those points, the force acting on the 2␲ 2␲ rad block has its maximum magnitude; compare Figs. 15-4a and The period follows from Eq. 15-2, which yields 15-4c, where you can see that the magnitudes of the dis- placement and acceleration are maximum at the same times. 1 1 T ϭ ϭ ϭ 0.64 s ϭ 640 ms. (Answer) f 1.56 Hz (e) What is the phase constant ␾ for the motion? (b) What is the amplitude of the oscillation? Calculations: Equation 15-3 gives the displacement of the block as a function of time. We know that at time t ϭ 0, ϭ KEY IDEA the block is located at x xm. Substituting these initial conditions, as they are called, into Eq. 15-3 and canceling xm With no friction involved, the mechanical energy of the spring– give us ϭ ␾ block system is conserved. 1 cos . (15-14) Taking the inverse cosine then yields Reasoning: The block is released from rest 11 cm from ␾ ϭ 0 rad. (Answer) its equilibrium position, with zero kinetic energy and the elastic potential energy of the system at a maximum. Thus, (Any angle that is an integer multiple of 2␲ rad also satisfies the block will have zero kinetic energy whenever it is Eq. 15-14; we chose the smallest angle.) again 11 cm from its equilibrium position, which means it (f) What is the displacement function x(t) for the will never be farther than 11 cm from that position. Its spring–block system? maximum displacement is 11 cm: ϭ Calculation: The function x(t) is given in general form by xm 11 cm. (Answer) Eq. 15-3. Substituting known quantities into that equation (c) What is the maximum speed vm of the oscillating block, gives us ϭ ␻ ϩ ␾ and where is the block when it has this speed? x(t) xm cos( t ) ϭ (0.11 m) cos[(9.8 rad/s)t ϩ 0] KEY IDEA ϭ 0.11 cos(9.8t), (Answer) ␻ The maximum speed vm is the velocity amplitude xm in Eq.15-6. where x is in meters and t is in seconds. Additional examples, video, and practice available at WileyPLUS 392 CHAPTER 15 OSCILLATIONS Sample Problem Finding SHM phase constant from displacement and velocity ϭ ␻ ␾ At t 0, the displacement x(0) of the block in a linear oscil- Calculations: We know and want and xm. If we divide lator like that of Fig. 15-5 is Ϫ8.50 cm. (Read x(0) as “x at Eq. 15-16 by Eq. 15-15, we eliminate one of those unknowns time zero.”) The block’s velocity v(0) then is Ϫ0.920 m/s, and reduce the other to a single trig function: and its acceleration a(0) is ϩ47.0 m/s2. Ϫ␻ ␾ ␻ v(0) ϭ xm sin ϭϪ␻ ␾ (a) What is the angular frequency of this system? ␾ tan . x(0) xm cos KEY IDEA Solving for tan ␾, we find With the block in SHM, Eqs. 15-3, 15-6, and 15-7 give its dis- v(0) Ϫ0.920 m/s tan ␾ ϭϪ ϭϪ placement, velocity, and acceleration, respectively, and each ␻x(0) (23.5 rad/s)(Ϫ0.0850 m) contains ␻. ϭϪ0.461. Calculations: Let’s substitute t ϭ 0 into each to see This equation has two solutions: whether we can solve any one of them for ␻.We find ␾ ϭϪ25° and ␾ ϭ 180° ϩ (Ϫ25°) ϭ 155°. ϭ ␾ x(0) xm cos , (15-15) Normally only the first solution here is displayed by a calcu- v(0) ϭϪ␻x sin ␾, (15-16) m lator, but it may not be the physically possible solution. To ϭϪ␻2 ␾ and a(0) xm cos . (15-17) choose the proper solution, we test them both by using them In Eq. 15-15, ␻ has disappeared. In Eqs. 15-16 and 15-17, we to compute values for the amplitude xm. From Eq. 15-15, we ␾ find that if ␾ ϭϪ25°, then know values for the left sides, but we do not know xm and . However, if we divide Eq. 15-17 by Eq. 15-15, we neatly elim- x(0) Ϫ0.0850 m ␾ ␻ x ϭ ϭ ϭϪ0.094 m. inate both xm and and can then solve for as m cos ␾ cos(Ϫ25Њ) a(0) 47.0 m/s2 ␾ ϭ ϭ ␻ ϭ Ϫ ϭ Ϫ We find similarly that if 155°, then xm 0.094 m. A x(0) A Ϫ0.0850 m Because the amplitude of SHM must be a positive constant, ϭ 23.5 rad/s. (Answer) the correct phase constant and amplitude here are ␾ ␾ ϭ ϭ ϭ (b) What are the phase constant and amplitude xm? 155° and xm 0.094 m 9.4 cm.
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