General Physics I

Spring 2011

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Oscillations

Oscillations• A quantity is said to exhibit oscillations if it varies with time

about an equilibrium or reference value in a repetitive

fashion.

• Oscillations are periodic when the time for one complete

cycle of the oscillations is constant. This repeat time is

called the period (T) of the oscillation.

• Let us assume that for a given oscillation, f cycles are

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• Let us assume that for a given oscillation, f cycles are

completed in one second. The number of cycles per

second is called the frequency of the oscillation. Since f

cycles are completed in one second, it follows that one

cycle is completed in 1/f seconds. The time taken to

complete one cycle is the period. Thus,

or 1 1.T fTf

= =

The unit of frequency is the hertz (Hz). 1 Hz = 1 cycle/sec.

Simple Harmonic Motion• Consider an object attached to

a spring at one end. The other

end of the spring is fixed. The

object rests on a frictionless

horizontal surface. The object is

in equilibrium when the spring is

not stretched or compressed. If

the object is displaced from the

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equilibrium position and

released, it will oscillate about

this position.

• If the displacement of the object

from its equilibrium position is

graphed as a function of time,

one obtains sinusoidal behavior

(sine or cosine function).

1/ .f T=

Simple Harmonic Motion

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Oscillatory motion for which the position is a sinusoidal function

of time is called simple harmonic motion.

Restoring Force• An oscillation must have a

restoring force. This is the net force

that always acts in a direction

toward the equilibrium position,

and thus tends to “restore” the

object back to the equilibrium

position. The oscillation is

sustained because when the object Marble in a bowl

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sustained because when the object

reaches the equilibrium position

(where the net force is zero), it has

a non-zero velocity and so its

inertia carries it past the

equilibrium position. The restoring

force then slows it down, it stops

momentarily and then moves back

toward the equilibrium position.

Marble in a bowl

Object attached to spring

Linear Restoring Force• The spring force is a linear restoring

force because the force is proportional to

the displacement from the equilibrium

position:

• The equation above is called Hooke’s

law. Note that the force is in the opposite

direction to the displacement of the free

( ) .sp xF k x=− ∆

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direction to the displacement of the free

end of the spring from its equilibrium

position. Thus, the force always points

toward the equilibrium position.

• If we take the equilibrium position to be x

= 0, then ∆x = x. So,

• A linear restoring is required for simple

harmonic motion.

( ) .sp xF kx=−

Slope is –k, where k is the

spring constant.

Workbook: Chapter 14, Questions 1, 2

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Describing Simple Harmonic Motion• The graph to the right shows the

position (or displacement) x versus time t for a particle undergoing simple harmonic motion (SHM). The position as time progresses is described by the cosine function:

2cos .tx ATπ

=

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In the equation above, A is the magnitude of the maximum displacement from equilibrium, which is the amplitude. When

x = A or –A, the particle is at its maximum distance from equilibrium and so the velocity is zero. The particle changes directions at these points, which are called turning points.

Describing Simple Harmonic Motion

• Note that the argument of the cosine function (the quantity inside the parentheses) must be in radians.

• At t = 0, the argument of the cosine function is zero. The cosine of zero radians = 1. Thus, x = A. The object starts at the maximum positive displacement from equilibrium.

• At t = T/4 (one-fourth of one period), the argument of the cosine function is π/2 radians. The cosine of π/2 radians is zero. Thus,

2cos .tx ATπ

=

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function is π/2 radians. The cosine of π/2 radians is zero. Thus, at this time, the object is at the equilibrium position.

• At t = T/2 the argument of the cosine function is π radians. The cosine of π radians is -1. Thus, at this time, the particle is at x = -A and is about to change directions.

• At t = 3T/4, the argument of the cosine function is 3π/2 radians. The cosine of 3π/4 radians is zero. Thus, the particle is at the equilibrium position again.

Describing Simple Harmonic Motion

• At t = T, the argument of the cosine function is 2π radians. The

cosine of 2π radians = 1. Thus, x = A. The object is back to its

starting point. One cycle has been completed.

Velocity

2cos .tx ATπ

=

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Velocity

• To find the velocity of the object at any instant, we can find the

slope of the position graph. It is easy to see that the slope is zero when x = ±A, so the velocity is zero at these times. These

are the turning points as mentioned before. Also, the slope has

maximum magnitude when x = 0, i.e., at the equilibrium. Thus,

the speed of the object is greatest at the equilibrium position.

Describing Simple Harmonic Motion• From the slope of the position

graph, we find that the velocity of

the object undergoing SHM is

given by

where

max2sin ,x

tv vTπ

=−

max2 .AvTπ=

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• Like the cosine function, the sine

function will be positive, zero, or

negative at different times. Thus,

the velocity will negative, zero, or

positive at these times. A negative

velocity simply indicates motion in

the negative x direction (leftward).

max .vT

=

Describing Simple Harmonic MotionAcceleration

• One can find the acceleration versus time by finding the slope

of the velocity vs. time graph. However, another way to obtain

the acceleration vs. time behavior is to use Newton’s second

law. The restoring force is the only force acting along the

direction of motion and so is the net force in this direction. The acceleration is given by ax =(Fnet)x/m. But, (Fnet)x = −kx. Thus,

k=−

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• We see that the acceleration is proportional to the position but

in the opposite direction to it. Thus, the acceleration vs. time

graph must also be a cosine graph, but an inverted one

because of the minus sign. Using our previous expression for x

as a function of t, we find that

.xka xm=−

2cos .xk kA ta xm m T

π

=− =−

Describing Simple Harmonic Motion• We can rewrite the acceleration

as a function of time as

where

max2cos ,x

taT

a π

=−

2

24 .max

AT

kAa mπ

==

13

T

Describing Simple Harmonic Motion

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A mass attached to a spring oscillates back and forth as

indicated in the position vs. time plot below. At point P, the

mass has

1. positive velocity and positive acceleration.

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1. positive velocity and positive acceleration.

2. positive velocity and negative acceleration.

3. positive velocity and zero acceleration.

4. negative velocity and positive acceleration.

5. negative velocity and negative acceleration.

6. negative velocity and zero acceleration.

7. zero velocity but is accelerating (positively or

negatively).

8. zero velocity and zero acceleration.

Vertical Springs• If an object is mounted on a

vertical spring, it will execute

simple harmonic motion in exactly

the same way as an object that is

attached to a horizontal spring.

• The vertical orientation causes

gravity to lower the equilibrium

position. That is all.

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position. That is all.

Workbook: Chapter 14, Question 7

Textbook: Chapter 14, Question 23,Problem 9

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