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Transcript
Note 12 Oscillatory Motion Sections Covered in the Text: Chapter 14, except 14.7 & 14.8 In this note we investigate the oscillatory motion of an object. Oscillatory motion is the motion an object executes as it moves back and forth through a position of stable equilibrium. We shall examine two examples of oscillatory motion, a mass on the end of a spring and the simple pendulum. Both stand as prototypes for oscillations of all kinds in nature, ranging from the movement of molecules of water on the surface of a pond to oscillations of electric and magnetic fields in electromagnetic radiation. Both examples are also featured in lab experiments most of you will be doing at about this time in the course. Motion of a Particle Attached to a Spring In Notes 08 and 09 we examined certain aspects of the physics of a mass on the end of a spring. But that study did not deal with the continuous motion of the mass. We consider the physics of that motion here. Consider an object (a block) modelled as a particle connected to the end of a spring and in contact with a horizontal frictionless surface (Figure 12-1).1 a function this force is F(x) = –kx . …[12-1] This function is called Hooke’s Law. Here k is a constant called the spring constant. The spring force is zero at the equilibrium position x = 0. We have seen in other notes that eq[12-1] is an example of a central or restoring force. It is restoring in the sense that if the block is displaced to the right of the equilibrium position, then the force exerted by the spring on the block points to the left, toward the equilibrium position. If the block is displaced to the left of the equilibrium position, then the force exerted by the spring on the block points to the right, again toward the equilibrium position. You should be able to see that if the block is displaced either right or left of the equilibrium position and then released, then the block will move back and forth through the equilibrium position under the action of the varying spring force—that is, the block will execute a vibrational or oscillatory motion. If the surface on which the block rests could be made to be frictionless, then no energy would be converted to thermal energy in the movement. In principle, the block would oscillate back and forth forever between positions equidistant on either side of the equilibrium position. This kind of motion is called simple harmonic motion (abbreviated SHM). The system executing such motion is called a simple harmonic oscillator. Mathematical Representation of SHM The motion just described in words can be put into mathematical form. Equating eq[12-1] to ma and rearranging we can obtain an expression for the block’s acceleration: Figure 12-1. A mass on the end of a spring is shown at three positions in the course of its oscillatory motion. The spring exerts a force on the block that is proportional to the spring’s displacement. Written as a=– k x. m …[12-2a] If the object is confined to 1D space then we can drop the vector notation and write eq[12-2a] as the function: a(x) = – k x. m …[12-2b] 1 This system might just as well be a glider on an airtrack connected between two identical springs. In this case the effective kvalue of the system is 2k. Now x is a function of time and a(x) can be written as the second derivative of x(t) with respect to time. Thus 12-1 Note 12 d 2 x(t) k = – x(t) . 2 dt m …[12-3] We shall see in what follows that we can write k/m as the square of a quantity ω: ω2 = k . m …[12-4] Eq[12-3] then becomes d 2 x(t) = –ω 2 x(t ) . 2 dt …[12-5] This equation is a second-order differential equation in x(t). A solution is 2 x(t) = Acos (ωt + φ ) ω= where k , m Figure 12-2. Graphs of eq[12-6] showing the meaning of the period T and amplitude A for arbitrary φ (a) and φ = 0 (b). …[12-6] a(t) = …[12-7] consistent with eq[12-4]. A is a constant. ω is called the angular frequency of the vibration. It has dimension € T–1 and units rad.s–1 (or just s–1 because rad is a dimensionless quantity). ω can also be written as ω = 2π f , …[12-8] dv(t) d 2 x(t) = = –A ω 2 cos(ωt + φ ) . 2 dt dt …[12-10] Note that v(t) and a(t) have the same sinusoidal form as x(t). Graphs of x(t), v(t) and a(t) for an arbitrary phase angle φ are shown in Figures 12-3. where f is the linear frequency. f has units cycles.s –1 or hertz (abbreviated Hz). φ is called the phase angle. Among other things, φ sets the value of x(t) at t = 0. The time for one complete vibration is called the period and is denoted T. T has dimension T and unit s. The period is the inverse of the frequency (T = f –1). The maximum displacement of the vibration from equilibrium is called the amplitude and is denoted A. Figure 12-2 shows a graph of eq[12-6]. Expressions for the velocity and acceleration of the block can be obtained from successive differentiations of eq[12-6] with respect to t. The results are: v(t) = and 2 dx(t ) = – Aω sin(ωt + φ ) dt …[12-9] To show that eq[12-6] is a solution of eq[12-5] you needn’t prove that it follows mathematically. It is sufficient to show that eq[12-6] along with its second derivative satisfy eq[12-5] by substitution. 12-2 Figure 12-3. Displacement, velocity and acceleration graphs of the vibration of a block on the end of a spring for an arbitrary phase angle φ. You should study Figures 12-3 to understand the relationships between the displacement, velocity and acceleration. For example, when the displacement is a Note 12 maximum positive value, the velocity is zero and the acceleration is a maximum negative value. Can you correlate these facts with the sketch of the motion in Figure 12-1? Let us consider a numerical example. K(t) = = 1 2 1 2 mv(t)2 , 1 2 mω 2 A2 sin 2 (ωt + φ ) = kA2 sin 2 (ωt + φ ) …[12-11] Example Problem 12-1 A Block-Spring System A block of mass 200. g is connected to a light horizontal spring of force constant 5.00 N.m–1 and is free to oscillate on a horizontal frictionless surface. (a) If the block is displaced 5.00 cm from equilibrium and then released from rest find the period of its motion. (b) what is the phase angle φ? Solution: (a) From eq[12-7] the period of motion of a mass on the end of a spring is given by ω= so that k 2π = 2πf = m T T = 2π € = 2π € using the expression for v(t) from eq[12-9]. The potential energy of the system (stored in the spring) has been seen (Note 08) to be given by 1 2 U( x) = kx2 , Substituting eqs[12-11] and [12-12] we have E = K(t) + U (t) = 1 2 1 2 kA2 sin 2 (ωt + φ ) + kA2 cos 2 (ωt + φ ) . This expression can be simplified. Collecting terms and using a trigonometric relation we have = 1.26 s. Note that€T is independent of amplitude. (b) According to the expression for the position of the block, eq[12-6], the position at time t = 0 is …[12-12] using eq[12-6]. The total mechanical energy E of the system is the sum of the kinetic and potential energies: E = K +U . m k 0.2(kg) 5.00(N.m−1 ) 1 2 U(t) = kA2 cos2 (ωt + φ ) so that E= 1 2 kA2 [sin2 (ωt + φ ) + cos 2 (ωt + φ)] , = 1 2 kA2 , …[12-13] x(0) = 0.05cos(ω0 + φ) = 0.05. Thus φ = 0. The value of the phase angle is set by the position of the block at t = 0. Energy Considerations in SHM It is instructive to study the mass on the end of a spring from the point of view of energy as well as from the point of view of kinematics. If the block is moving with velocity v(t) then it possesses kinetic energy K(t) given by since the expression in square brackets is unity. The total energy is therefore proportional to the square of the amplitude. Remarkably, E, which is equal to the sum of two time dependent functions, is itself independent of time. K and U can be written as functions of either x or t. They are graphed as functions of t in Figure 12-5a and of x in Figure 12-5b. Clearly, at the instant when the kinetic energy is a maximum the potential energy is a minimum and vice versa. In one period the kinetic and potential energies go through two oscillations. Both functions are sinusoidal. Notice that at x = 0, the potential energy is zero and the kinetic energy is a maximum (since v is a maximum). At x = ±A, the potential energy is a maximum and the kinetic energy is zero (since v is zero). At all 12-3 Note 12 positions of x the sum of K and U is constant, just as at all times t the sum of K and U is constant. Be prepared to describe a vibrating system such as this one in words. The Simple Pendulum A simple pendulum is essentially an object of mass m suspended on the end of a string (Figure 12-6).3 For the simple pendulum to work as intended, gravity must be present. The system has an equilibrium position defined when the mass is in a rest position with the string vertical. In this position the angular displacement of the string with respect to the vertical is θ = 0 (rad) or the linear displacement of the mass relative to the rest position along the arc of its path is s = 0 (m). Figure 12-6. A simple pendulum showing the forces on the mass and their components. Figure 12-5. Graphs of K and U as functions of time (a) and of x (b). For convenience, the phase angle in (a) is taken to be zero. When the mass is displaced so that the string makes a non-zero angle with respect to the vertical and released, the pendulum will swing back and forth. If friction with the air is negligible, then the mass will swing continuously between maximum angular positions θmax and –θmax forever. This kind of motion is simple harmonic motion. The pendulum system is a simple harmonic oscillator. As in the case of a mass on the end of a spring, this motion can be described mathematically. In the case of a mass/spring system the driving force is the spring force. In the case of a simple pendulum the driving force is the component of the force of gravity, –mgsinθ, acting tangent to the particle’s path. Applying the second law F = ma we get –mg sin θ (t) = m 3 d 2 s(t) , dt 2 …[12-14] The story goes that Galileo studied the motion of a simple pendulum by observing the swinging motion of a chandelier in the local cathedral. Lacking high technology he counted time in units of beats of his pulse. 12-4 Note 12 where s(t) is the position of the particle measured with respect to the equilibrium position along the particle’s path. Substituting s(t) = Lθ(t), where L is the length of the string, eq[12-14] can be written d 2 θ (t) g = – sin θ (t) . 2 dt L …[12-15] This is a second order differential equation in θ (t). Eq[12-15], unlike eq[12-3], is very difficult to solve. However, if we assume that the angular displacement θ(t) is small, then we can make the substitution sinθ(t) ≅ θ(t). Then eq[12-15] reduces to d 2 θ (t) g = – θ (t) . 2 dt L …[12-16] Apart from a change in variable and the presence of different constants, eq[12-16] is the same as eq[12-3]. Furthermore if we put ω= g , L …[12-17] then eq[12-16] becomes of the same form as eq[12-5]. It must therefore have the same form of solution, namely, € θ (t) = θ max cos(ωt + φ ) . …[12-18] The function θ (t) is sinusoidal as is the function x(t) (Figure 12-6). The period of the motion is thus given by T= 2π L = 2π . ω g The mathematics for a simple pendulum is thus seen to be the same in essentials as for a mass on the end of a spring. This underlines a major goal of physics, to describe various systems with the same tools whenever possible. A critical difference between the two systems is that the period of a simple pendulum involves g whereas the period of the motion of a mass on the end of a spring does not. A mass on the end of a spring will function as expected in a spacecraft just as well as on Earth, but a simple pendulum will not work in a spacecraft. Example Problem 12-2 The Simple Pendulum A simple pendulum (unlike a mass on the end of a spring) can be used as an instrument to calculate the local acceleration due to gravity—if the period can be measured with good accuracy. Assuming the period of a simple pendulum of length 1.00 m is measured to be 2.00709 s what is the local value of g? Solution: Rearranging eq[12-19] we have for g: g= 4π 2 L 4π 2 1.00m = T2 (2.007s) 2 = 9.80 m.s–2, to 3 significant figures. …[12-19] € 12-5 Note 12 To Be Mastered • • Definitions: angular frequency, linear frequency, period, phase angle General expression for the position of a simple harmonic oscillator: x(t) = Acos(ωt + φ ) • • • Physics of: mass on the end of a spring Physics of: the simple pendulum Difference between the period of a mass on the end of a spring and the period of a simple pendulum (memorized): T = 2π • m k T = 2π L g Sketches of K, U and E for both systems € € Typical Quiz/Test/Exam Questions 1. Define the following and give their units: (a) angular displacement (b) angular velocity (c) angular frequency 2. A simple pendulum is studied at some position on the surface of the earth. If the length of the pendulum is increased by a factor of 4 then the period of the pendulum (a) decreases by a factor of 2 (b) descreases by a factor of 4 (c) increases by a factor of 2 (d) remains unchanged. 3. The period of a simple pendulum of a fixed length L is measured at the surface of the Earth and at the top of a high mountain. At the surface of the Earth the period of the pendulum is To and the acceleration due to gravity is go. If at the top of the mountain the period is 2To then what is the acceleration due to gravity there in terms of go? 4. The position of a block on the end of a spring (as shown in Figure 12-1) is given by the function x(t) = 2.0 cos(3.0t + π / 2) meter. The constant of the spring is 9.0 N.m –1. Assume that friction between block and surface is negligible. Answer the following questions: (a) What is the maximum displacement of the block (in m) from the equilibrium position? (b) What is the angular frequency (in rad.s–1) of the motion? (c) What is the mass of the block (in kg)? (d) What was the position of the block at a time t = 0? 12-6 Note 12 5. A simple pendulum consists of a bob of mass m on the end of a string of length l suspended from a stationary support (see figure). When released from near position A at time t = 0, the bob swings through B to near C and back to A repetitively. Assume that friction of the air is negligible. Answer the following questions. support θ l m A B C (a) What kind of motion does the bob execute and why? (b) Sketch one complete cycle of the angular position of the pendulum starting from t = 0. Show clearly on the sketch the period and amplitude. (c) Sketch the kinetic and potential energy of the bob over one complete cycle starting from t = 0. (d) What is the angular acceleration of the bob at position B? 6. A glider of mass 400.0 g is placed on a horizontal airtrack and connected between two springs whose opposite ends are connected to block supports that cannot move (see the figure). Each spring has a k value of 1.17 N.m –1. The equilibrium position of the glider (x = 0) is as shown. The glider is displaced 5.0 cm from equilibrium and then released. Friction is negligible. Answer the following questions. block block glider airtrack x=0 (a) Calculate the maximum speed of the glider in m.s–1. (b) Calculate the speed of the glider in m.s –1 when it is 3.0 cm from equilibrium. (c) Derive an expression for the acceleration a of the glider in terms of its instantaneous position x. (d) What is the acceleration of the glider in m.s–2 when it passes through x = 0? Reminder: The effective k-value of this two spring system is 2k. 12-7 Note 12 7. A block is connected to the end of a spring and supported on a horizontal frictionless surface. The block is released at a clock time of t = 0 s from a position x = +A m (see the figure). Answer the following questions. (a) Write down an expression for the force exerted on the block by the spring. (b) What kind of motion does the block execute and why? (c) What is the total work done on the block by the spring on one complete cycle of the motion? (d) What is the acceleration of the block at the equilibrium position? 8. A glider released from the upper end of a curved air track moves back and forth through the lowest point on the track with simple harmonic motion (see the figure). Answer the following questions: glider C A B (a) Draw a free body diagram of the glider at positions A, B and C. (b) What is the total work done by the gravitational field of the Earth on one cycle of the glider from A back to A? 12-8