Lecture-X

Small oscillation

Stability:The result F = -dU/dx is useful not only for computing the force but also for visualizing

the stability of a system from the potential energy plot.

Suppose there is a force on the particle is F = -dU/dx, and the system is in equilibrium

where there is no force i.e. dU/dx = 0. If this occurs at a minimum of U it is a stable

equilibrium whereas if it is at a maximum of U, the equilibrium is unstable. Say, dU/dx

= 0 occurs at some point xo. To test for stability we must determine whether U has a

minimum or a maximum at xo. One needs to examine d2U/dx2 at xo.

If the second derivative is positive, the equilibrium is stable; if it is negative, the system

is unstable. If d2U/dx2 = 0, one must look at higher derivatives. If all derivatives vanish

so that U is constant in a region about xo, the system is said to be in a condition of

neutral.

Energy and Stability-The Teeter ToyThe teeter toy consists of two identical weights which hang from a peg on drooping

arms, as shown. Find the condition for stability of the toy with respect to rocking.

The solution is θ =0, as we expect from symmetry.

At equilibrium,

For the second derivative to be positive,

Bounded and unbounded motion:Since kinetic energy K=E-U can never be negative, the motion

of the system is constrained to regions where U < E.

Harmonic oscillator: total energy is constant, E is

represented by a horizontal line. Motion is limited to the

shaded region where E > U; the limits of the motion, x1

and x2, are called the turning points. The motion is

bounded.

U=A/r: There is a distance of closest approach, rmin, but

the motion is not bounded for large r since U decreasesmin

the motion is not bounded for large r since U decreases

with distance. Unbounded motion.

A commonly used potential energy function to describe

the interaction between two atoms is the Lennard-Jones

potential

For E > 0, the motion is unbounded, and the atoms are free to fly apart.

For E < 0, the motion is bounded, and the atoms never approach closer

than ra or move farther apart than rb.

Time period of bounded motion

( )

( )

2

1

21

,2

2 or

2

x

x

dxK m E U

dt

E Udx dxt

dt m E U

m

= = −

−= =

−∫

Since,

where x1 and x2 are the turning points. The time period is then T=2t.

b b

U=2U0

U=U0

U=0

E=3U0/2

Consider the potential shown in the figure. Find the period of oscillation.

( )

2

0 00 0

0

0

32 2

11

3

1Thus, 2 1

3

b b

b

dx dx m mt b b

U UE E U

m m

mb

U

mT b

U

= + = +−

= +

= +

∫ ∫

Small oscillationNearly every bound system oscillates like a harmonic oscillator if it is slightly perturbed

from its equilibrium position.

Expand U(r) about r0, the position of the potential

minimum.

where the effective spring constant.

0

2

2

2

1( ) , 0,

r

U k d Umx F x kx x x

x m m drω ω

∂= = − = − + = = =

∂�� ��

Frequency of small oscillation:

Molecular VibrationsSuppose that two atoms of masses m1 and m2 are bound together in a molecule with energy so

low that their separation is always close to the equilibrium value ro. With the parabola

approximation, the effective spring constant is k=(d2U/dr2)Iro. How can we find the vibration

frequency of the molecule?

This vibrational motion, characteristic of all molecules,

can be identified by the light the molecule radiates. The

vibrational frequencies typically lie in the near infrared

(3X1013 Hz), and by measuring the frequency, the value of

d2U/dr2 at the potential energy minimum. For the HCI

molecule, the effective spring constant turns out to be

500 N/m.

Frequency from energy pre-factors:

If the potential and kinetic energies are given by:

The total energy can be obtained as:

where q is a generalized displacement.

Since the total energy is constant:

The equation of motion can be obtained as:

A

Bω =where2

0q qω+ =��or

Example: Small oscillation of teeter toy

2 2 22 coss L l Ll α= + −

Hence,

Formal solution of Simple harmonic oscillator: Consider a spring mass system:

If at time t = 0 the position of the mass is x(0) and

its velocity v(0),

The amplitude A and the phase φ are given by

The time period of motion

Energy Considerations

Let us calculate the time average values of the potential and kinetic energies over one

For a harmonic oscillator:

and

&

The total energy is constant as expected.

Let us calculate the time average values of the potential and kinetic energies over one

period of oscillation

Thus,

The time average kinetic and potential energies are equal. When friction is present,

this is no longer exactly true.

Damped harmonic oscillation:

Consider a spring mass system with a damping force

Net force acting on the system is

Equation of motion:

/ ,b mγ =Or, with

tx x e

α=Trial solution: 0

tx x e

α=Trial solution:

1 2t t

A Bx x e x e

α α= +

where xA and xB are constants.

Then the general solution can be written as

The motion is similar to the undamped case except

that the amplitude decreases exponentially in time

and the frequency of oscillation ω1 is less than the

undamped frequency ω0 .

( ) ( )2

1cos

tx Ae t

γ ω φ−= +Then,

Consider

From the work energy theorem:

Physically, E(t) decreases with time because the friction force continually dissipates

energy.

How E(t) depends on time can be found by calculating the kinetic and potential

energies K(t) and U(t). The velocity v is given by

If the motion is only lightly damped, 'Y / WI « 1, and the coefficient of the second term

in the bracket is small. Then and

and

( ) ( )2

1cos

tx Ae t

γ ω φ−= +

Since the damping is assumed to be small,

and taking

, ,

The energy decreases exponentially in time.

The Q of an Oscillator

The degree of damping of an oscillator is often specified by a dimensionless parameter

Q, the quality factor, defined by

The energy dissipated in a short time ∆t is

The time to oscillate through one radian isThe time to oscillate through one radian is

Hence, one radian of oscillation requires time

A lightly damped oscillator has Q »1.