Lecture 18 – Oscillations about

Equilibrium

Periodic Motion

Period: time required for one cycle of periodic motion

Frequency: number of oscillations per unit time

This unit is called the Hertz:

Simple Harmonic MotionA spring exerts a restoring force that is proportional to the displacement from equilibrium:

Displaced, at rest

Moving, past equilibrium point

Displaced, at rest

Moving, past equilibrium point

Displaced, at rest

This is called “Simple Harmonic Motion”

Simple Harmonic Motion

A mass on a spring has a displacement as a function of time that is a sine or cosine curve:

Here, A is called the amplitude of the motion.

Simple Harmonic MotionIf we call the period of the motion T – this is the time to complete one full cycle – we can write the position as a function of time:

The position at time t +T is the same as the position at time t, as we would expect.

Time Positiont =0 x = A

t=Tx = Acos(2π) = A

t = T/2

x = Acos(π) = -A

t = T/4

x = Acos(π/2) = 0

T

Sine vs Cosine

x at t=0 : A

x at t=0 : 0

v at t=0 : 0

v at t=0 : >0

Harmonic Motion IHarmonic Motion I

a) 0a) 0

b) b) AA/2/2

c) c) AA

d) 2d) 2AA

e) 4e) 4AA

A mass on a spring in SHM has

amplitude A and period T. What

is the total distance traveled by

the mass after a time interval T?

Harmonic Motion IHarmonic Motion I

a) 0a) 0

b) b) AA/2/2

c) c) AA

d) 2d) 2AA

e) 4e) 4AA

A mass on a spring in SHM has

amplitude A and period T. What

is the total distance traveled by

the mass after a time interval T?

In the time interval time interval TT (the period), the mass goes

through one complete oscillationcomplete oscillation back to the starting

point. The distance it covers is The distance it covers is A + A + A + AA + A + A + A = =

(4(4AA).).

A mass on a spring in SHM has amplitude A and period T. What is the net displacement of the mass after a time interval T?

a) 0

b) A/2

c) A

d) 2A

e) 4A

Harmonic Motion IIHarmonic Motion II

A mass on a spring in SHM has amplitude A and period T. What is the net displacement of the mass after a time interval T?

a) 0

b) A/2

c) A

d) 2A

e) 4A

The displacement is x = x2 – x1. Because the

initial and final positions of the mass are the same (it ends up back at its original position), then the displacement is zero.

Harmonic Motion IIHarmonic Motion II

Follow-upFollow-up: What is the net displacement after a half of a period?: What is the net displacement after a half of a period?

The Pendulum

A simple pendulum consists of a mass m (of negligible size) suspended by a string or rod of length L (and negligible mass).

The angle it makes with the vertical varies with time as a sine or cosine.

How a pendulum is like the mass on a spring

Looking at the forces on the pendulum bob, we see that the restoring force is proportional to sin θ

The restoring force for a spring is proportional to the displacement

This is the condition for simple harmonic motion

Approximation for sin θ

However, for small angles, sin θ and θ are approximately equal.

θ (deg) θ (rad) sin(θ)

1 0.01745 0.01745

5 0.08727 0.08716

10 0.1745 0.1736

20 0.3491 0.3420

Pendulum for small angles = simple harmonicfor small angles of the

pendulum bob, the restoring force is proportional to θ

F = -mg θ = -mg s / L

The restoring force for a spring is proportional to the displacement

F= -kx

So: the motion of the angle of the pendulum is the same as the motion for the mass on a spring, with k mg/L

Uniform Circular Motion and Simple Harmonic Motion

An object in simple harmonic motion has the same motion as one component of an object in uniform circular motion

Assume oscillation of mass on the spring has the same period T as the circular motion of the peg on the “record player”

Here, the object in circular motion has an angular speed of:

where T is the period of motion of the object in simple harmonic motion.

The position as a function of time:

... just like the simple harmonic motion!

Position of Peg in Circular Motion

Velocity of Peg in Circular Motion

Linear speed v = Aω

x component:

Acceleration of Peg in Circular Motion

Linear acceleration a = Aω2

x component:

Summary of Simple Harmonic MotionThe position as a function of time:

The angular frequency:

From this comparison with circular motion, we can see:

The velocity as a function of time:

The acceleration as a function of time:

Speed and AccelerationSpeed and Acceleration

a) x = A

b) x > 0 but x < A

c) x = 0

d) x < 0

e) none of the above

A mass on a spring in SHM has

amplitude A and period T. At

what point in the motion is v =

0 and a = 0 simultaneously?

Speed and AccelerationSpeed and Acceleration

a) x = A

b) x > 0 but x < A

c) x = 0

d) x < 0

e) none of the above

A mass on a spring in SHM has

amplitude A and period T. At

what point in the motion is v =

0 and a = 0 simultaneously?

If bothIf both vv andand aa were zero were zero

at the same time, the mass at the same time, the mass

would be at rest and stay would be at rest and stay

at rest!at rest! Thus, there is NONO

pointpoint at which both vv and aa

are both zero at the same

time.Follow-upFollow-up: Where is acceleration a maximum?: Where is acceleration a maximum?

The Period of a Mass on a SpringFor the mass on a spring:

Substituting the time dependencies of a and x gives

and the period is:

The Period of a Mass on a Spring

Vertical SpringWhat if the mass hangs from a vertical spring?

Fs=kx

dx=0

W=mg

new equilibrium position: x= -d = -mg/k

total force as a function of x:

x=-d

with

Looks like the same spring, with a different equilibrium position (x’=0 -> x = -d)

Simple harmonic motion is unchanged from the horizontal case!

Energy Conservation in Oscillatory Motion

In an ideal system with no nonconservative forces, the total mechanical energy is conserved. For a mass on a spring:

Since we know the position and velocity as functions of time, we can find the maximum kinetic and potential energies:

Energy Conservation in Oscillatory Motion

As a function of time,

So the total energy is constant; as the kinetic energy increases, the potential energy decreases, and vice versa.

0

Period of a Pendulum

A pendulum is like the mass on a spring, with k=mg/L

Therefore, we find that the period of a pendulum depends only on the length of the string:

Physical Pendula

A physical pendulum is a solid mass that oscillates around its center of mass, but cannot be modeled as a point mass suspended by a massless string. Examples:

Period of a Physical Pendulum

In this case, it can be shown that the period depends on the moment of inertia:

Substituting the moment of inertia of a point mass a distance l from the axis of rotation gives, as expected:

Damped Oscillations

In most physical situations, there is a nonconservative force of some sort, which will tend to decrease the amplitude of the oscillation, and which is typically proportional to the speed:

This causes the amplitude to decrease exponentially with time:

Damped Oscillations

This exponential decrease is shown in the figure:

“underdamped” means that there is more than one oscillation

Damped OscillationsThe previous image shows a system that is underdamped – it goes through multiple oscillations before coming to rest.

A critically damped system is one that relaxes back to the equilibrium position without oscillating and in minimum time;

an overdamped system will also not oscillate but is damped so heavily that it takes longer to reach equilibrium.

Driven Oscillations and ResonanceAn oscillation can be driven by an oscillating driving force; the frequency of the driving force may or may not be the same as the natural frequency of the system.

Driven Oscillations and Resonance

If the driving frequency is close to the natural frequency, the amplitude can become quite large, especially if the damping is small. This is called resonance.

A mass on a spring oscillates in simple harmonic motion with amplitude A. If the mass is doubled, but the amplitude is not changed, what will happen to the total energy of the system?

a) total energy will increase

b) total energy will not change

c) total energy will decrease

Energy in SHM IEnergy in SHM I

A mass on a spring oscillates in simple harmonic motion with amplitude A. If the mass is doubled, but the amplitude is not changed, what will happen to the total energy of the system?

a) total energy will increase

b) total energy will not change

c) total energy will decrease

The total energy is equal to the initial value of the elastic potential energy, which is PEs = kA2. This

does not depend on mass, so a change in mass will not affect the energy of the system.

Energy in SHM IEnergy in SHM I

Follow-upFollow-up: What happens if you double the amplitude?: What happens if you double the amplitude?

A mass oscillates on a vertical spring with period T. If the whole setup is taken to the Moon, how does the period change?

a) period will increase

b) period will not change

c) period will decrease

Spring on the MoonSpring on the Moon

A mass oscillates on a vertical spring with period T. If the whole setup is taken to the Moon, how does the period change?

a) period will increase

b) period will not change

c) period will decrease

The period of simple harmonic motion depends only on the mass and the spring constant and does not depend on the acceleration due to gravity. By going to the Moon, the value of g has been reduced, but that does not affect the period of the oscillating mass–spring system.

Spring on the MoonSpring on the Moon