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 with f(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 = MgQM = Mg(PS-PR) PS=rθcosθ PR=asinθ

I parallel axis = IG + Md2 Parallel Axis Theorem For Cube of sides 2a : IG= ⅔ Ma2 d → QG = [a2 + (rθ)2]½ ≈ a (for small angle) IQ = ⅔Ma2 + 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