Force and dynamics with a spring, analytic approach

Transcription

1 Force and dynaics with a spring, analytic approach It ay strie you as strange that the first force we will discuss will be that of a spring. It is not one of the four Universal forces and we don t use springs every day. Or do we? The basic otion of a spring is to oscillate. Many, any echanical systes oscillate. However, they ight not appear to be springs. Tae a tree branch bouncing up and down in the wind, is it a spring? How about the vibrations between two atos in a olecule or a cor bobbing up and down on the surface of water, are these spring systes? Fro a physics point of view, all of these can be odeled by the sae equations as those that describe a spring. This is why exaining spring systes is so iportant. Later in the class we will ae this atheatically precise, but for now just reeber that the study of springs is far, far ore general and iportant than a coil of etal with a ass hanging fro it. The force law for an ideal spring is F ( L L0 ) L. This says that the force that a spring applies is proportional to the aount that the spring is stretched fro its equilibriu length. The constant of proportionality is negative, eaning that the force is directed in the opposite direction fro the extensive or copressive displaceent L. Let s loo at the siplest spring syste: L 0 L 0 L L F In this figure the ass is displaced to the right and the spring exerts a restoring force to the left. dp Writing Newton s equation for otion in the lateral direction gives x ( x L0 ), dt dx() t d x() t cobining with vt () and nonrelativistically p( t) v( t) gives ( x L 0). dt dt This is a linear second-order inhoogeneous differential equation with constant coefficients (that is a outh-full), with a haronic solution. We have turned the physical proble of a spring, through Newton s equations of otion, into a atheatical proble: solving a differential equation. Now we need to solve the equation to learn about the behavior of the ass-spring syste. There are a nuber of ways to solve a differential equation; one is siply guessing the solution. This is perfectly acceptable, but often d x() t x L 0 difficult. The equation before us, ( ), is quite siple. We can guess one dt solution right away, x() t L0. This is a correct solution, however pretty boring. The ass is

2 siply sitting at the equilibriu position of the spring and not oving. To find ore interesting solutions we need to consider the equation that reoves that particular solution: d x() t () xt. This equation says that after taing two derivatives of the function you get dt the function bac ultiplied by a constant. This leads us to consider a function that when differentiated reains proportional to itself: naely the exponential x() t Ae t. Rewriting the equation as d x t () ( ) 0 xt and substituting in the exponential gives 0. This is dt called the secular equation. The solutions are. Now, this is an interesting situation because both and are positive, but let s proceed. So we have two independent solutions for i t the proble x() t Ae and x() t Ae i t. The general solution is the su of these, plus i t i t the particular solution x() t L0. Specifically the general solution is x() t Ae Ae L0. This equation has two adjustable constants A and A. These can be deterined by using the initial conditions. So, we have analytically solved the atheatical proble. The only issue is that the solution is coplex and certainly not the siple oscillatory otion that we expected. Let s start by renaing the quantity. has the units of inverse tie and is called the angular frequency. We need to understand what the coplex exponential do this is expand the function in a Taylor series it e it t i t t i t t i t t...! 3! 4! 5! 6! 7! 8! i t e eans. The best way to This infinite series does not loo uch better. It is still coplex and contain an infinite nuber of ters. Let s siplify a bit by collecting real and iaginary ters: it e t t t t... it t t t...! 4! 6! 8! 3! 5! 7! n t t n it e i n! (n)! n0 n0 or

3 Again, this ight not see to iprove the situation until you realize that those two suations are the Taylor series expansions for the cosine and sine function. Go ahead and try it yourself, expand the cosine and sine function and find that n0 t cos( t) t t t t...! 4! 6! 8! n! and n0 n t sin( t) t t t t.... 3! 5! 7! n! This siplifies things a bit and produces one of the ost aazing equations in atheatics, it naely Eulers relation: e cos( t) isin( t). Now this is progress: we have produced the coon oscillatory functions that we expected. However, it is still a coplex expression. That it can be fixed. Loo at the siilar expression for the other solution e cos( t) isin( t), here we have used the fact that cosine is an even function, and sine is odd. By adding these two it it e e solutions together we get cos( t). This is still a solution to the original proble. e Siilarly subtraction yields sin( t) it e i it n. Having now done all of this we have i t i t refored our original coplex solution x() t Ae Ae L0 into a new for x( t) Asin( t) Bcos( t) L0 that contain only real functions ( A and B could still be coplex if we wanted, typically we don t). Often the solution is written in a second for that contains a phase angle. This change is siply a trigonoetric identity. x( t) Asin t Bcos t L0 or x( t) Csin t L0 These two expressions for xt () are equivalent in that there are two free paraeters in each AB, in the first and C, in the second and the function is periodic with frequency f. The two free paraeters can be deterined fro two additional conditions such as the initial position and initial velocity.

4 There is a relationship between the paraeters in each expression so that you can go fro one to the other as suits your needs: B C A B and tan, alternately A C cos and B C sin. This can be A reebered fro basic trigonoetry, see the figure. A C A B For now we will use x( t) Ccos t L0 The constant L0 is a reflection of where the origin is placed. If it is placed at the equilibriu location of the end of the spring, L0 will be zero. This is a coon way to set up your coordinate syste. The aplitude of oscillation is given by C, and the range of y varies fro C to C. The ter inside the parenthesizes, t, is called the phase. There is a tie independent ter, called the phase shift,initial phase, or phase constant, that deterines where in the cycle the otion starts at tie zero. In other words, it deterines whether the object starts at the equilibriu position and has a nonzero oentu or at a point of axiu displaceent with zero oentu, or anywhere in between these positions. To deterine C and you ust be given two pieces of inforation about the syste, such as fro what position did it start and how fast was it going initially. This will allow you to uniquely deterine these constants. So what about the tie dependent part of the phase, t? This ter deterines the frequency of oscillation. Every tie this part of the phase increases by the otion has copleted one

5 ore full cycle. That eans that the period T for an oscillation is given by T or T. If that is the tie per cycle, the nuber of cycles per second (Hz), the frequency f, can be found fro f T. A closely related quantity, the angular frequency, gives the oscillation rate in radians per second and is defined by f. Much of the behavior of the oscillator can be understood fro this figure: C xt () () t C where () t t. Picture a vector rotating with a constant rate, i.e. the second hand on a cloc, only this rotates counter-clocwise. The position of the oscillator is given by the x- coponent of the vector, i.e. where the dashed line on the figure intersects the x-axis. The vector rotates at a rate radians per second and the position of the oscillator ranges fro C to C. This is called haronic otion. On a position vs. tie plot the otion loos lie a sine curve:

6 3 The paraeters that were used are C,, 4 The progra at the end of this section has an aniation with a rotating vector and the position of the y-coordinate tracing out oscillations of haronic otion.

7 # Plots sine curve and rotating vector representation for haronic otion fro future iport division fro visual iport * fro visual.graph iport * # iport graphing features sine_function = gdisplay( title = 'haronic position vs. tie', xtitle = 'tie', ytitle = 'position', x=0, y=400, width=000, foreground=color.blac, bacground= color.white) #set display # sine_func = gdots(gdisplay = sine_function, color = color.blac) # curve for sine dt =. #graphing increent C = # C is aplitude of oscillation w = # angular frequency phi = 3*3.4/4 # phase shift, set initial conditions #graphics vector = arrow(pos=(0,0,0), axis=(c*cos(phi),c*sin(phi),0), shaftwidth=.0) position = sphere(pos = (C*cos(phi),0,0), radius =.05, color = color.red) for t in arange (0, 0, dt): # tie points fro 0 to 0 interval dt rate(0) sine = C*cos(w*t + phi) # haronic function sine_func.plot (pos=(t, sine )) # plot haronic function vector.axis = (C*cos(w*t + phi),c*sin(w*t + phi),0) # rotating arrow position.pos = (C*cos(w*t + phi),0,0) # x position of arrow