Oscillations

4a. The Simple Harmonic Oscillator

In general an oscillating system with sinusoidal time In general, an oscillating system with sinusoidal time dependence is called a harmonic oscillator. Many physical systems have this time dependence: mechanical oscillators, elastic systems, AC electric circuits, sound vibrations, etc.

In particle mechanics, the simplest harmonic oscillator consists of a mass attached to a spring, moving without friction. The figure shows a block with mass m attached to one end of a spring and p gmoving without friction on a horizontal floor; the other end of the spring is attached to an immovable wall.

We’ll solve the dynamics of the block, assuming the spring obeys Hooke’s law.

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Spring ForcesRobert Hooke, a contemporary of Isaac Newton (*), Robert Hooke, a contemporary of Isaac Newton ( ), found that spring forces can be described by some simple properties …

The spring has an equilibrium length.

If stretched or compressed by a small displacement, x, a restoring force pulls or pushes the spring toward equilibrium length.

Within the elastic limit, the force is linear in the displacement; F(x) = −k x.

(*) Hooke and Newton were acquaintances but their relationship was not friendly

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but their relationship was not friendly.

Dynamics of a mass on a spring

The motion of the block is an example of single-particle dynamics and one dimensional motion.

The equation of motion is

Or,

kxxFxm −== )(&&

kxx =−= ωω where2&&

It is well-known that the solutions of this equation are sinusoidal: sin ωt or cos ωt or, most generally, a linear combination

m

x(t) = A cos ωt + B sin ωt

The constants, A and B, must be determined from initial conditions or some other information about the oscillator. They cannot be determined from the differential equation!

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Example 1. General initial conditions.

Suppose we are given initial conditions,

x(0) = x and v(0) = vx(0) = x0 and v(0) = v0.

The general solution is

x(t) = A cos ωt + B sin ωt

v(t) = −ω A sin ωt + ω B cos ωt

The initial conditions are

x(0) = A = x0 and v(0) = ω B = v0.

Thus the solution is

x(t) = x0 cos ωt + (v0/ω) sin ωt

Example 2. Suppose the initial position is x = 0, and the amplitude of oscillator is R. Then what is x(t)?

After a little thought you’ll see that the solution is x(t) = R sin ωt.

Example 3. The mass is 1 kg and Hooke’s p gconstant is 100 N/cm. What is the frequency of oscillation?

1-s 7.1521

2 Frequency ===

mk

fππ

ω

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22 mππ

Example 4. EnergyConsider the general solution

x(t) = A cos ωt + B sin ωt(A) Calculate the kinetic and potential energy.(B) Calculate the time averages of K and U.(C) Calculate E and verify that it is constant.(D) Calculate the amplitude of oscillation.

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Epilogue (4a)

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..

than

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Oscillations

4b – The Damped Oscillator

Resistance force; viscosity

Restoring force; Hooke’s law

SummaryResults in terms of the original parameters

⎨⎧ =

xbkxxmb/m2β

&&&⎩⎨ =

−−=mk

xbkxxm/

20ω

Over-d d b2 > 4mk strong

d i β > ω0damped b 4mk damping β > ω0

Critically damped b2 = 4mk optimal

damping β = ω0Damped Oscillator

Under-damped b2 < 4mk weak

damping β < ω0

Damped Oscillator

0.8

1.0

0.0

0.2

0.4

0.6

x(t)

-0.4

-0.20.00 0.50 1.00 1.50 2.00 2.50

time

for ω0=1.0 and β=0.02

0.000

0.500

1.000

0 0 10 0 20 0 30 0 40 0 50 0 60 0 70 0 80 0 90 0 100 0

for ω0 1.0 and β 0.02

-1.000

-0.5000.0 10.0 20.0 30.0 40.0 50.0 60.0 70.0 80.0 90.0 100.0

0.500

1.000

-1.000

-0.500

0.0000.0 10.0 20.0 30.0 40.0 50.0 60.0 70.0 80.0 90.0 100.0

Note that v = 0 at ω1 t = π, 2π, 3π, 4π, ...These are the maximum displacements from x = 0.ese a e t e a u d sp ace e ts o 0The maximum positive displacements occur at

ω1 t = 0, 2π, 4π, 6π, …i.e., ω1 tn = 2π n for n = 0, 1, 2, 3, 4 …

xmax,n = A exp(−β tn) = A exp (−2π β n/ω1)