Oscillations Different Situations Oscillations Free Oscillations Forced Oscillations Under NO damping
(Undamped Oscilations) Under damping (Damped Oscillations) x(t)=Real part of z(t)= x+iy
Undamped Free Oscillations x(t)=Real part of z(t)= x+iy Damped Free Oscillations
Resistive force is proportional to velocity Undamped Forced Oscillations Damped Free Oscillations Linear Differential Equations
and it Solutions homogeneous inhomogeneous
Linear differential equation of order n=2 homogeneous or inhomogeneous Complementary Solution
Take trial solution : x = emx, m is constant m1, m2,.will be the roots If all roots are real and distinct, then the general solution: x =c1em1t+c2em2t+. If some roots are complex, if a+ib is one then a-ib will be the other, solution: x = eat(c1 cos(bt) +c2 sin(bx)) + If some roots are repeated, say m1 repeated k times, then the solution:x = (c1 + c2t+ ..cktk-1)em1t General solution = Complimentary + Particular solution
Particular Solution & General Solution General solution = Complimentary + Particular solution For the inhomogeneous one, the complementary solution is obtained in the same way ,i.e., by making f(t)=0 Then the particular solution is added: the solution to be assumed depending on the form of f(t) 2nd Order Linear Homogeneous Diff. Eqn
2nd order linear homogeneous differential equation with constant coefficients General solution : x1 (t) and x2 (t) are linearly independent, i.e x1 (t) NOT proportional to x2 (t) 2nd Order Linear Inhomogeneous Diff. Eqn
2nd order linear inhomogeneous differential equation with constant coefficients General solution : Complementary function Particular integral: obtained by special methods, solves the equation withf(t)0; without any additional parameters A & B : obtained from initial conditions Free Oscillations: SHO
A simple harmonic oscillator is an oscillating system which satisfies the following properties 1. Motion is about an equilibrium position at which point no net force acts on the system 2. The restoring force is proportional to and oppositely directed to the displacement 3. Motion is periodic This is the first slide Elements of an Oscillator
need inertia, or its equivalent mass, for linear motion moment of inertia, for rotational motion inductance, e.g., for electrical circuit need a displacement, or its equivalent amplitude (position, voltage, pressure, etc.) need a negative feedback to counter inertia displacement-dependent restoring force:spring,gravity, etc. electrical potential restoring charges Harmonic Oscillator Potential Model System Consider a mass m attached to one end of a light (massless) helical spring whose other end is fixed. We choose the origin x=0 where spring is unstretched. The mass is in stable equilibrium at this position and it will continue to remain there if left at rest We are interested in a situation where the mass is disturbed from equillibrium The mass experiences a restoring force from the spring if it is either stretched or compressed. Simple Pendulum Torsional Oscillation Compound Pendulum Electrical Oscillations Solution A=Amplitude, =Phase we see the equation of motion is
Solution: Form-1 We wish to find x(t) : we see the equation of motion is the obvious solution is of the form of sine or cosine function x(t) is a function describing the oscillation what function gives itself back after twice differentiated, with negative constant? cos(at), sin(at) both do work exp(at) looks like it ought to work... Uniform Circular Motion vs SHO So the general solutions is
Solution: Form-1 So the general solutions is where A1 and A2 are determined by the values of x and dx/dt at a specified time So the general solution is given by the addition or superposition of both values of x so we have Solution: Form-2 If we rewrite the constants as Where is a constant angle, then So that And finally Exponential Solution: Form-3
A=Complex amplitude Real and imaginary parts of z(t) satisfy simple harmonic equation of motion x(t)=Re z(t) Solution in Three Forms Another Example The Problem Cube of mass M, side 2a
Cylinder of radius r, fixed along a horizontal axis Cube rocks on the cylinder for small angles, does not slip Find time period Torque about Q = MgQM = Mg(PS-PR) PS=rcos PR=asin I parallel axis = IG + Md2 Parallel Axis Theorem
For Cube of sides 2a : IG= Ma2 d QG = [a2 + (r)2] a (for small angle) IQ = Ma2 + Ma2 = 5/3Ma2 @AR Equation of Motion Torque about Q: Moment of inertia of the cube about a horizontal axis passing through Q is I I = Using small angle approximation, Energy of SHO K.E. & P.E. also vibrates Phase space