Physics 430 Lecture 11 Oscillations Dale E Gary NJIT Physics Department 5 1 Hooke s Law As you know Hooke s Law spring law gives the force due to a spring in the form assuming it is along the x axis Fx x kx where x is the displacement relative to the spring s equilibrium point The force can be in either direction in the x direction when x is negative and in the x direction when x is positive at least for the case when k is positive Since the force is conservative we can write the force as minus the gradient of a potential energy U x 12 kx 2 From what we have learned about potential energy graphs you can immediately see that this upward curving parabola indicates that x 0 is a stable equilibrium if k were negative the parabola would s be curved downward and x 0 would be an unstable equilibrium However Hooke s Law has a general validity for the following reason for any general potential energy curve U x in the vicinity of an equilibrium point at x xo which we can take to be zero we can always perform a Taylor series expansion U x about U 0 that U 0point x 12 U 0 x 2 October 7 2010 Hooke s Law 2 The first term is the constant term but since potential energy can be defined with any zero point the constant term can be ignored or considered to be zero The second term is the linear term but since x 0 is an equilibrium point the slope there is by definition zero so the first non zero term is the third term U x 12 kx 2 This says that for sufficiently small displacements from an equilibrium point Hooke s Law is ALWAYS valid for any potential energy function This justifies our consideration of this case in detail Of course Hooke s Law can relate to any coordinate not just x Let s revisit the box on a cylinder problem as an illustration October 7 2010 Example 5 1 Cube Balanced on a Cylinder Statement of the problem Consider again the cube of Example 4 7 and show that for small angles the potential energy takes the Hooke s Law form U k 2 Solution We found before that the potential energy for the cube in terms of was U mg r b cos r sin For small we can make the approximations cos 1 2 2 sin Notice that we kept the TWO leading terms for cos Why With these substitutions we have U mg r b 1 12 2 r 2 mg r b 12 mg r b 2 Ignoring the constant term we see that this is in the form of Hooke s Law with a spring constant k mg r b Notice that the equilibrium is stable only when k 0 i e when r b as we found after quite a bit more work before Notice also that since is a parabola the turning points at 2 1 U x 2 kx are equidistant from the equilibrium least for small displacements point October 7 2010 5 2 Simple Harmonic Motion Let s now look at all of this from the point of view of the equation of motion Consider a cart on a frictionless track attached to a spring with spring constant k Since we have Fx x kx the equation of motion is k x x 2 x m x kx m where we introduce the constant x k x 0 m which represents the angular frequency with which the cart will oscillate as we will see Because Hooke s Law always applies near equilibrium for any potential energy we will find oscillations to be very common governed by the general equation of motion such as for a 2 pendulum vs angle October 7 2010 Exponential Solutions The equation x 2 x is a second order linear homogeneous differential equation Therefore it has two independent solutions which can be written x t e i t and x t e i t As you can easily check both of these solutions do satisfy the equation However you should have come to expect two arbitrary constants in the solution to a second order differential equation two constants of integration and in fact any linear combination of these x t C1e i t C2 e i t two solutions is also a solution This is called the superposition principle which works for any linear system Any solution containing two arbitrary constants is in fact the general solution to the equation e i t can cos twritten i sin in t terms of Sine and Cosine by using This solution be Euler s equation Plugging these x t C t i C t can B1write cos t B2 sin t into our original solution we 1 C 2 cos 1 C2 sin October 7 2010 Sine and Cosine Solutions Thus the solutionsx t C1e i t C2 e i t are equivalent so long as and x t B1 cos t B2 sin t B1 C1 C2 and B2 i C1 C2 An important issue is that x t being a actual position coordinate has to be real In general the first solution looks as if it is complex while the second solution looks as if it is real However this depends on whether the coefficients are real or not Both C1 and C2 can be complex but if both their sum and difference C1 C2 and C1 C2 are real then B1 and B2 are real and in fact both versions are completely equivalent This solution is the definition of simple harmonic motion SHM As usual we determine these constants from the initial conditions If I start the motion by pulling the cart aside position x 0 xo and releasing it v 0 0 then only cosine x t the xo cos t term survives and vo If the cart starts at x 0 0xby giving itt a kick velocity v 0 vo then t sin only the sine term survives and October 7 2010 Phase Shifted Cosine Solution The two pure solutions just described are shown in the figures below x xo x x t xo cos t vo t x t vo sin t t slope v o x slope vo o The general solution if I both pull the cart to the side and give it a push both xo and vo non zero the motion is harder to visualize but you may expect it to be the same type of motion with some phase shift We can demonstrate that for the general case by defining another 2 2 A B1 B2 constant A B2 where A is the hypotenuse of a right triangle whose sides are B1 x t B cos t B sin t B1 B2A cos Bthere A sin B1 and Obviously is also an associated angle such 1 2 2 A cos cos t sin sin t that Thus Note that you haveand already solved this same problem using energy in problem 4 28 A cos t October …