Created by a disturbance in a mediumMedium must vibrate back and forth at a specific periodUsually involves the motion of a mass on a springCould be horizontal or verticalMass initially rests at the equilibrium positionMoved to a maximum displacement (compression or extension) known as AMPLITUDE

F is force in NewtonsNegative b/c force is in opposite direction of displacementk is spring constant in N/m higher values imply stiffer springx is displacement from initial position in meters

Spring will attempt to return to equilibrium position by exerting a restoring force F = - kx

System will oscillate with a period (T) such that T = 2 (m / k) ½

The system will have frequency f = 1 /T

Energy stored in an oscillator will be Ep = ½ kx2, at max amplitude Ep = ½ k A2 and all energy is potentialAs mass moves from max amp toward equilibrium point Ep Ek

Giancoli uses vo at equilib, I recommend vmax

At equilib pt. All energy is Ek = ½ mvmax2 specific to Equilib pt: * ½ kA 2 = ½ mvmax

2

AT ANY GIVEN POINT IN THE OSCILLATION: ** ½ kA2 = ½ mv2 + ½ kx2

Simple Harmonic Motion (SHM)

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Recall that when undergoing circular motion the object passes through a distance of one circumference in exactly one period g iving v = 2r / T

where v is vmax. Solve for T gives T = 2r / vmax but if the radius is A then this becomes T = 2A / vmax

Now solving * for vmax gives vmax = A k / m this is substituted into the above giving T = 2A / A k / m

As A factors out and the fraction in the denominator is inverted we get:

† T = 2 m / k

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We can use the reference circle to examine the distance from the center in the x direction.

Notice that the object starts at max distance A and from the triangle x = A cos

At any time t the angle can be found in radians as = 2 t / T as the full circle 2 is completed in one period T. This shows that any x value is a function of time by subbing in for q giving:

††x = A cos (2t / T)

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††x = A cos (2t / T)

dx = d [A cos (2t / T)] = A d [ cos (2t / T)] x d [2t / T] = - A sin (2t / T) x 2 / Tdt dt dt dt

v = A [ - sin (2t / T)] x [2 / T] = - 2A / T sin (2t / T) Since : T = 2A / vmax we get :

††† v = - vmax sin (2t / T)

The velocity v at any moment can be found as the derivative x with respect to t

ma = – kA cos (2t / T)

The acceleration may be found using Newton’s 2d law as a = F /m or a = – kx / m and subbing from †† yields:

Not really a harmonic oscillatorPretty straight forward

Pendulums: T = 2 L / g

Oscillations decrease in amplitude over time due to energy “loss” from resistance in systemOver damped: takes a long time to reach equilibUnder damped: over shoots equilibCritically damped: fastest possible route to equilib and still a function (see fig. 11 – 17 page 321)

Damped harmonic motion::

Vibrations near fo will cause the object to vibrate itself with wave amplitude increasing to the breaking point due to superposition.Resonance: vibrations of an object have a natural frequency fo = 1 / T = 1 / 2 k / m

Wave eqn: v = f, in a string v = Tension / (mass / length)

Pasted from <file:///D:\My%20Documents\My%20Files\strachan\Physics%2012%20AP\SHM\Simple%20Harmonic%20Motion%20notes.doc>

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