Simple Harmonic Motion

Transcription

1 Simple Harmonic Motion (FIZ 101E - Summer 2018) July 29, 2018 Contents 1 Introduction 2 2 The Spring-Mass System 2 3 The Energy in SHM 5 4 The Simple Pendulum 6 5 The Physical Pendulum 8 6 The Damped Oscillations 9 7 Forced Oscillations & Resonance 12 1

2 1 Introduction In this part of the course we consider a special type of motion called periodic motion, the repeating motion of an object in which it continues to return to a given position after a xed time interval. The repetitive movements of such an object are called oscillations. We will focus our attention on a special case of periodic motion called simple harmonic motion. The universal importance of SHM is that to a good approximation many real oscillating systems behave like simple harmonic oscillators when they undergo oscillations of small amplitude. Consequently, the elegant mathematical description of the simple harmonic oscillator that we will develop can be applied to a wide range of physical systems. 2 The Spring-Mass System Oscillatory behavior, or harmonic motion, can be encountered in diversely many physical systems; but to understand the fundamentals of this special type of motion, the prototype system that we consider is the spring-mass system. First we will consider only the springs obeying Hooke's Law : F spring = k x = kxî (1) where k is the spring constant (a characteristic property of a string for its stifness) and x is the amount of displacement from the equilibrium position x 0 = 0, as shown in the gure: Forces tending to bring the system back to the (stable) equilibrium point are called restoring forces, and we always encounter oscillatory motion when there is a restoring force. In this part of the course, we will analyze linear restoring forces in detail, and see how to treat the case of non-linear restoring forces by a linear approximation. 2

3 The Newton's Second Law for such a system gives and the equation of motion is F net = m a, (2) kxî = m d2 x î, dt2 (3) (m d2 x + kx)î dt2 = 0, (4) d 2 x(t) + k x(t) = 0 (5) dt 2 m This is one of the most important equations of physics, called Simple Harmonic Motion Equation (SHM). Note that it is a 2nd order dierential equation. Linear restoring forces lead to SHM. Actually it is not clear at this step that why this equation is called SHM equation, or how it implies harmonic or periodic motion. It is the solution of this dierential equation which leads to this naming. Solving this equation means that we should get x(t) in terms of the given parameters of this dierential equation. The form of this dierential equation leads us to guess the following forms of the solution: x(t) = A cos(ωt + φ) or x(t) = A sin(ωt + φ) or x(t) = A cos(ωt) + B sin(ωt). (6) Note that in all three solutions there are three unknown parameters: (A, ω, φ) or (A, B, ω). But from now on just as a consensus I'll use the form x(t) = A cos(ωt + φ). If we directly substitute this guessed solution to the Eq.(5), we get ω = k m (7) which is called as the angular frequency. Recall that period, T, is the time for one cycle in a periodic/harmonic motion, and frequency, f, is the total number of cycles in a second, and f = 1/T. From the form of the solution, if T is the period, we expect ωt = 2π ω = 2π T = 2πf (8). This means that for the spring-mass system, m T = 2π k or f = 1 k 2π m. (9) What about the other parameters A and φ? They are determined by the so-called initial conditions; x(0), v(0), etc. (For a second order dierential equation we need two initial conditions to determine the exact form of the solution.) 3

4 Having determined the form of the solution x(t) which gives the displacement of the mass from the equilibrium position as a function of time, we can determine the speed and acceleration of the mass also as a function of time: v(t) = dx dt = ωa sin(ωt + φ) v max = ωa = k A, (10) m a(t) = d2 x dt = 2 ω2 A cos(ωt + φ) a max = ω 2 A = k A. (11) m Example A block of mass m making SHM along the x-axis is at the origin at t = 0 and moving along the +x-direction. If the amplitude of the SHM is 2 cm and the frequency of oscillations is 1.5 Hz. (a) Find the displacement function x(t). (b) Determine v max and a max. (c) Determine the total distance traveled between t = 0 and t = 1 s. (d) At which instant is it at x = 3 cm? 4

5 3 The Energy in SHM The 1-dimensional forces of the form are all conservative forces. Then using the relation we can get the potential energy function for SHM: F (x) = ±f(x)î (12) F (x) du î, (13) dx F (x) = kxî U(x) = 1 2 kx2, U(0) = 0. (14) The total mechanical energy of a system executing SHM is (ignoring friction) E = K + U = 1 2 mv2 (t) kx2 (t) = E = 1 2 ka2 : constant. (15) It is also important to see an explicit example of the connection between conservation laws and equations of motion in the spring-mass system: d {E = 12 dt mv } kx2 d2 x dt + k x = 0. (16) 2 m Actually it is sometimes easier to get the equation of motion in this way. Example Consider a spring-mass system on a frictionless, horizontal surface; assume that the spring does not obey the Hooke's Law but F (x) = cx 3ˆx where c is a constant. The amplitude of the oscillations is A. (a) What is the physical dimension and the SI unit of c? (b) Calculate the potential energy function U(x) of this spring, taking U(0) = 0. (c) The quarter period is the time required to move from x = 0 to x = A. Calculate the period of the oscillations in this system. (d) If the speed of the block at the equilibrium position is v 0 = 2 m/s, nd the speed of the block when it is at x = A/2. (e) Is this motion SH? Explain. 5

6 4 The Simple Pendulum A simple pendulum is an idealized model consisting of a point mass suspended by a massless and unstretchable string, as shown in the gure: In order to obtain the equation of motion for this system, it is better to use quantities which are appropriate for rotational motion; thus, instead of considering F net = m a, we consider τ net = Iα with respect to the point of suspension, O. The torque with respect to point O is called the restoring torque, and the equation of motion is, then, and this nally becomes, τ net,o = I o α mg sin θl = ml 2 d2 θ dt, (17) ( ) 2 d 2 θ mgl dt + sin θ = 0, (18) 2 ml 2 θ + ( g L) sin θ = 0 (19) This equation describes a kind of harmonic motion but not SHM! But noting that the Taylor expansion of the sin θ is sin θ = θ 1 3! θ ! θ5, (20) this becomes for small oscillations sin θ θ, and the Eq(19) becomes, ( g θ + θ 0 (21) L) which is the (approximate) equation of SHM. The solution is, recalling the spring-mass system, θ(t) = θ max cos(ωt + φ) where ω 2 = g L and T = 2π L g. (22) 6

7 How small is really small for this approximation to be valid? The exact expression for the period of a pendulum without any approximation is L T = 2π { g 2 2 sin2 ( θ max 2 ) sin4 ( θ } max 2 ) + (23) 7

8 5 The Physical Pendulum A physical pendulum is an extended object that, for small angular displacements, can be modeled to move in simple harmonic motion about a pivot that does not go through the center of mass, as shown in the gure: The equation of motion for this system is, by considering τ net,o = I o α again as in the case of simple pendulum, θ + mgd sin θ = 0, (24) I o and in the small angle approximation, the equation of motion becomes ( ) mgd θ + θ 0, (25) I o and this is now the equation of motion for SHM; in this equation I o = I cm + md 2 because of the parallel-axis theorem, and the angular frequency of the oscillations is mgd ω =. (26) I o Example Consider the uniform rod of (M, L), pinned to a wall at its midpoint in the vertical plane. A point mass of m is attached to the upper end and a massles spring is attached to the lower end (of the force constant k). (For a uniform rod of (M, L) I km = 1 12 ML2.) (a) Calculate the exact equation of motion for the oscillations of the system. (b) Calculate the approximate equation of motion for small oscillations of the system, and calculate the period of small oscillations. 8

9 6 The Damped Oscillations Almost all physical systems are aected by friction or resistive (drag) forces. These forces tend to remove energy from the moving system and thereby slow it down (i.e. damp the motion): Thus, there is an energy loss due to damping and such a decrease in energy in turn leads to a decrease in amplitude. The type of the drag force that we will consider in this section is the so-called velocitydependent drag force of the form F D = b v = bvˆv where b is a constant called the damping parameter. This type of force is very successful in modelling many systems involving viscous uids, such as the motion in air, in shock absorbers, etc., and the force can be generalized experimentally as F D = f(v)ˆv. When the mass is displaced from its equilibrium position there will be the restoring force due to the spring and in addition the damping force bvˆv due to the uid. The resulting equation of motion is, F net = m a ma = kx bv, (27) and this becomes ( ) ( ) d 2 x b dx k dt + 2 m dt + x = 0. (28) m This is the equation of a damped harmonic oscillator and it is a second order dierential equation; the damping force is linearly proportional to velocity and this linear dependence is very convenient as it has led to an equation that we can readily solve. A damping force proportional to, say v 2 would be much more dicult to handle. Fortunately, this linear dependence is a good approximation for many other oscillating systems when the velocity is small. The educated guess for the solution is of the form x(t) = Ae αt cos(ωt + φ), (29) and this represents a solution decreasing in amplitude - note that you can also consider the factors in front of the cosine function as the decreasing amplitude A(t) = Ae αt. The 9

10 parameters A and φ are determined from the initial conditions and the other two parameters are obtained by directly substituting this solution to the Eq(28): α = b 2m and ω2 = k m b2 4m 2 = k m α2 = ω 2 0 α 2. (30) The ω 0 is considered as the natural frequency of the oscillations. The form of the oscillation frequency leads to the existence of dierent solutions depending on the degree of damping involved: 1. Underdamped Oscillations : ω > 0, and it exhibits oscillatory behavior, albeit with decreasing ammplitude, and the solution is of the form x(t) = Ae αt cos(ωt + φ). 2. Critically Damped Oscillations : ω = 0 b c = 4km, and the system no longer oscillates but returns to its equilibrium position without oscillation when it is displaced and released, and the form of the solution is x(t) = Ae αt cos φ. 3. Overdamped Oscillations : b > b c ω is imaginary; again there is no oscillation, but the system returns to equilibrium more slowly than with critical damping, and the solution is of the form x(t) = Ae a 1t + Be a 2t. 10

11 The mechanical energy is not conserved in damped oscillations, as expected. The rate of energy loss in damped oscillations can be calculated as follow: de dt = d { 1 dt 2 mv2 + 1 } 2 kx2 = v(ma + kx) = v( bv), (31) and this can be written as The generalization of this result to other forms of the damping is de dt = bv2 (32) de dt = F D v (33) Obviously the energy decresaes but not at a uniform rate. (In electrical circuits, recall that, P = i 2 R; the reistance plays the role of damping parameter b.) Example A 10.6 kg object oscillates at the end of a vertical spring that has a spring constant of k = N/m. The eect of air resistance is represented by the damping coecient b = 3.00 Ns/m. (a) Calculate the frequency of the damped oscillation. (b) By what percentage does the amplitude of the oscillation decrease in each cycle? (c) Find the time interval that elapses while the energy of the system drops to 5% of its initial value. 11

12 7 Forced Oscillations & Resonance A damped oscillator left to itself will eventually stop moving altogether. But we can maintain a constant amplitude oscillation by applying a force that varies with time in a periodic way, with a denite period and frequency. This additional force is called driving force. A common example of a forced oscillator is a damped oscillator driven by an external force that varies periodically, such as F (t) = F 0 sin(ωt), where F 0 is a constant and ω is the angular frequency of the driving force. In general, the frequency ω of the driving force k is variable, whereas the natural frequency ω 0 = is xed. Modeling an oscillator with m both retarding and driving forces as a particle under a net force, Newton's second law in this situation gives m d2 x dt + bdx 2 dt + kx = F 0 sin(ωt). (34) Again, the solution of this equation is rather lengthy and will not be presented, but we will try to get it on physical arguments. After a suciently long period of time, when the energy input per cycle from the driving force equals the amount of mechanical energy lost (transformed to internal energy for each cycle), a steady-state condition is reached in which the oscillations proceed with constant amplitude, and the frequency of the oscillations is determined by the driving frequency, i.e. the solution is of the form x(t) = A cos(ωt + φ), (35) where A = F 0 /m (ω 2 ω0) ( ). (36) bω 2 m 12

13 For small damping, the amplitude is large when the frequency of the driving force is near the natural frequency of oscillation, or when ω ω 0. The dramatic increase in amplitude near the natural frequency is called resonance, and the natural frequency ω 0 is also called the resonance frequency of the system. Example A 2.00 kg object attached to a spring moves without friction (b = 0) and is driven by an external force given by the expression F (t) = 3.00 sin(2πt) where F is in newtons and t is in seconds. The force constant of the spring is k = 20.0 N/m. Find (a) the resonance angular frequency of the system once the steady-state is reached, (b) the angular frequency of the driven system, (c) the amplitude of the motion. 13