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SIMPLE HARMOIC MOTION CCHS Physics QuickTime™ and a TIFF (Uncompressed) decompressor are needed to see this picture

# SIMPLE HARMOIC MOTION CCHS Physics. Facts of SHM SHM occurs when an object is vibrating at a single frequency and period PERIODOIC: when a vibration or

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SIMPLE HARMOIC MOTION

CCHS PhysicsQuickTime™ and a

TIFF (Uncompressed) decompressorare needed to see this picture.

Facts of SHM

• SHM occurs when an object is vibrating at a single frequency and period

• PERIODOIC: when a vibration or oscillation repeats itself, back and forth, over the same path

• We’ll deal mainly with simple harmonic oscillations where the position of the object can be described by a sinusoidal (sin, cos) function

Examples of SHM

• Block attached to spring

• Motion of a pendulum

• Vibrations of a stringed musical instrument

• Clocks

• Oscillations of houses, bridges, …

Meaning ofSimple Harmonic Motion

• SIMPLE single frequency

• HARMONIC sinusoidal

• A system in which the restoring force is proportional to the negative displacement (Hooke’s Law) will move in simple harmonic motion.

Spring and SHM

Example of Complex Harmonic Motion

Approximately simple-harmonic in this region

Hooke’s Law• The magnitude of the restoring force is

proportional to the displacement

• Where:– F = force (N)– k = spring constant (N/m)– x = displacement

• Acceleration

• Force and Acceleration– Proportional to x– Directed toward the equilibrium position

F =−kx

a =−km

x

Hooke’s Law cont.

• Accurate as long as there is not too much displacement

• There is a negative sign because the force always acts opposite to the displacement

• Note: F varies with position

Five Fun Definitions

• DISPLACEMENT: the distance x from the equilibrium point

• AMPLITUDE: the maximum displacement• CYCLE: one complete to-and-from motion• PERIOD (T): time to complete one cycle• FREQUENCY (f): number of cycles per

second– Frequency is measured in s-1 = Hertz (Hz)

Period and Frequency

• Period and Frequency are inversely related:

f =1T

T =1f

Force vs. Distance Graphs

• Slope = spring constant• Area = work (or energy)

QuickTime™ and aTIFF (Uncompressed) decompressor

are needed to see this picture.

Area =12

bh

=12

x( ) kx( )

Work=12

kx2 =Energy

PEspring =Us =12

kx2

kxdx0

x

∫ =12

kx2

Energy of a Spring

• Remember that potential energy is stored energy due to position

• The calculation of potential energy is equivalent to calculating work

• Note: usually going to neglect mass of spring

PEspring =Us =12

kx2

Example 1Given the following graph:• What is the spring constant?

• If a 2 kg mass attached to this spring is displaced 3 m, what is the acceleration of the mass?

• If the spring is stretched to 5 m, how much energy is stored in the spring?

0

20

0 5

disp (m)

Force (N)

slope=205

k =4 N/m

a =kxm

=4 3( )2

a =6 m/s2

Us =12

kx2 =12

4( ) 5( )2 U =50 J

Conservation of Energy

• Mechanical Energy - the sum of all kinetic and potential energies– Now we add in elastic potential energy

• We still have that energy must be conserved, thus:

Etotal =KE +Ug +Us

Etotal =12

mv2 + mgh+12

kx2

1

2mv1

2 + mgh1 +12

kx12 +WNC =

12

mv22 + mgh2 +

12

kx22

Einitial =E final

Example 2• A spring stretches by 5 cm when a 100 g

mass is hung from it. A 250 g mass attached to this spring and pulled back 0.70 m and released from rest. What is the speed of the block as it passes

through the equilibrium point? • First, find the spring constant:

• Now, use conservation of energy

F =kx

.1 9.8( ) =k .05( )⇒ k = 19.6 N/m

1

2mv1

2 + mgh1 +12

kx12 +WNC =

12

mv22 + mgh2 +

12

kx22

1

2kx1

2 =12

mv22

1

219.6( ) .7( )2 =

12

.250( )v22

v =6.19 m/s

Example 3• A 0.2-kg ball is attached to a vertical massless spring as

shown. The spring constant of the spring is 28 N/m. The ball, supported initially so that the spring is neither stretched nor compressed, is released from rest. In the absence of air resistance, how far does the ball fall before being brought to a momentary stop by the spring?

• Conservation of Energy

– Note: x2 = h1 call this distance d

1

2mv1

2 + mgh1 +12

kx12 +WNC =

12

mv22 + mgh2 +

12

kx22

mgh1 =12

kx22

.2 9.8( )d=12

28( )d2

d =0.14 m

Velocity and Position

• Initially the mass is at its maximum extension A– Entirely elastic potential energy

• The initial elastic potential energy converted to a combination of kinetic and eleastic potential energy

• Solving for v

1

2kA2 =

12

mv22 +

12

kx22

v =±km

A2 −x2( )

Simple Harmonic Motion vs. Uniform Circular Motion

• A ball is glued to the top of a turntable• Focus on the shadow of the ball• If the turntable rotates with constant angular velocity the

shadow of the ball moves in simple harmonic motion

SHM vs. UCM cont.• If we can prove that the velocity of the shadow varies with position

like the function on the previous slides, we know the shadow moves in SHM. v =C A2 −x2

• Looking at the top triangle (made with velocity vectors)

• Now looking at the larger triangle

• Equating these two equations

sinθ =vv0

sinθ =A2 −x2

A

v

v0

=A2 −x2

A

v =v0A

A2 −x2 =C A2 −x2

• Thus we have proved that the velocity of the shadow in the x direction is related to displacement in exactly the same manner as the velocity of an object undergoing SHM.

Frequency and Period• The velocity of the ball in the previous slide is:

or• Where

– A is the amplitude– T is the period

• Focus on 1/4 of the trip,• Imagine the shadow is a block on a spring, in the 1/4

of a cycle the block moves from a point of all elastic potential energy to a point of all kinetic energy

• Substituting for A/v0

v0 =2πAT

T

4=2πA4v0

T =2πAv0

1

2kA2 =

12

mv02 ⇒

A

v0

=m

k

Ts =2πmk

Period of a Pendulum• Restoring force = mgsinθ• Not SHM because F sinθ, not θ• However, for small θ:

sinθ ≈ θ (measured in radians)• Thus, F = -mgsinθ ≈ mgθ• Also, s = Lθ

• Essentially SHM (Hooke’s Law)• Plugging into period of a spring

formula with k = (mg)/L

Ft =−mgL

⎛⎝⎜

⎞⎠⎟s

T =2πmk

=2πm

mg L Tp =2πLg

Displacement Equation

• Looking at reference circle again • Ball starts on the x axis at x = +A and

moves through the angle θ in a time t• Ball rotates at constant angular speed

(because UCM) θ = t• Displacement x of the shadow is just

the projection of the radius A onto the x-axis

x =Acosθ =Acos t( )

x(t) =Acos(t+φ)

…Angular frequency

A…amplitude

…phase constant, phase angle

Position of particle at time t:

)cos()( tAtx =

T…period

t …phase

Properties of SHM

Period and Frequency (Revisited)• PERIOD: time required to complete one cycle• The value of T depends on the angular speed

(frequency) – Greater , less T

• For 1 cycle, θ = 2π, and t = T:• Using the fact that frequency and period are

inverses:

=θt

=angular displacement

time =

T

=2π f

=2π

T= 2π f

Velocity and Acceleration

v =−Asin(t+φ)

a =− 2Acos(t+φ)

x =Acos(t+φ)

note: v t( ) =dxdt

and a t( ) =dvdt

Ax =max

maxv A=

2maxa A=

Phase of velocity differs by – π/2 or 90° from phase of displacement.

Phase of acceleration differs by – π or 180° from phase of displacement.

More Properties of SHM

• Acceleration of particle is proportional to the displacement, but is in the opposite direction (a = - 2·x).

• Displacement, velocity and acceleration vary sinusoidally.

• The frequency and period of the motion are independent of the amplitude.

Even More Properties of SHM

t x v a KE Us

0 A 0 -2A 0 ½kA2

T/4 0 -A 0 ½kA2 0

T/2 -A 0 -2A 0 ½kA2

3T/4 0 -A 0 ½kA2 0

T A 0 -2A 0 ½kA2

Summary of SHM

ExampleA block whose mass m is 680 g is fastened to a spring whose spring constant k is 65 N/m. The block is pulled a distance of x = 11 cm from its equilibrium position at x = 0 on a frictionless surface and released from rest at t = 0.

a) What force does the spring exert on the block just before the block is released?

b) What are the period, frequency and angular frequency?

F =−kx=− 65( ) .11( ) = −7.2N

Ts =2πmk

=2π.68065

= 0.64 s

=2π f = 2π 1.56( ) = 9.78 rad/s

f =1T

=1.64

=1.56 Hz

Example cont.c) What is the amplitude of the oscillation?

The amplitude is 11 cm (that is where it is released from)

d) What is the maximum speed of the oscillating block?

max speed occurs whenever x = 0

e) What is the magnitude of the maximum acceleration of the block?

max acceleration occurs at the end points

vmax =A=9.78 .11( ) =1.1 m/s

amax = 2A=9.782 .11( ) =11 m/s2

Example cont.f) What is the phase constant of the motion?

At t = 0, the displacement is at its maximum value A, and the velocity of the block is zero. If we put these initial conditions into the displacement and velocity equations we find:

The smallest angle that solves these two equations is = 0. (Any multiple of 2π will also work)

x =Acos(t+φ)

.11=.11cos 0( ) +φ( )

1=cos φ( )

v =−Asin(t+φ)

0 =1.1sin 0( ) +φ( )

0 =sin φ( )