Contributed by:

Oscillatory or Periodic motion, Phase diagrams, Energy conservation in oscillatory motion, Damped oscillations, Driven oscillations, and resonance

1.
What are the similarities and

differences between graphs?

Mass on spring

Both show oscillatory or cyclic

motion

differences between graphs?

Mass on spring

Both show oscillatory or cyclic

motion

2.
Oscillatory/Periodic Motion

Repetitive Motion with a period and frequency.

• Motion can be driven internally - mass on spring.

• Externally – pendulum or tides.

Repetitive Motion with a period and frequency.

• Motion can be driven internally - mass on spring.

• Externally – pendulum or tides.

3.
Guitar string

EEG Beta Waves

EEG Beta Waves

4.
SHM – special case

• What do you notice?

• What do you notice?

5.
Vocabulary

• Period T: time required for one cycle of

periodic motion (sec).

• Frequency: number of oscillations per unit

time.

unit is Hertz:

• Period T: time required for one cycle of

periodic motion (sec).

• Frequency: number of oscillations per unit

time.

unit is Hertz:

6.
What are the period and frequency of the

ECG

• T = 1 sec.

• f = 1 hz,

ECG

• T = 1 sec.

• f = 1 hz,

7.
• Angular Frequency - (rad/sec)

= 2f.

• remember angular speed from circular motion?

= /t

• = 2 rad/s

• Equilibrium (0): the spot the mass would come to rest

when not disturbed – Fnet = 0.

• Displacement: (s or x): distance from equilibrium.

• Amplitude (xo) – max displacement from eq.

= 2f.

• remember angular speed from circular motion?

= /t

• = 2 rad/s

• Equilibrium (0): the spot the mass would come to rest

when not disturbed – Fnet = 0.

• Displacement: (s or x): distance from equilibrium.

• Amplitude (xo) – max displacement from eq.

8.
Representation of Oscillatory Motion

• Observe the motion of a bobbing mass. Where is the:

• Displacement positive?

• Displacement negative?

• Displacement zero?

• velocity positive?

• Where is the velocity negative?

• Where is the velocity zero?

• Observe the motion of a bobbing mass. Where is the:

• Displacement positive?

• Displacement negative?

• Displacement zero?

• velocity positive?

• Where is the velocity negative?

• Where is the velocity zero?

9.
Simple Harmonic Motion – SHM

• Isochronous period.

• Restoring Force directly proportional to displacement. Double displacement, double

force etc.

• Further from equilibrium, more force directed toward it.

• Force and displacement in opposite directions.

• Isochronous period.

• Restoring Force directly proportional to displacement. Double displacement, double

force etc.

• Further from equilibrium, more force directed toward it.

• Force and displacement in opposite directions.

10.
Pendulum is not SHM

• Fnet not directly opposite s

• for small displacement angles it approximates SHM.

• Fnet not directly opposite s

• for small displacement angles it approximates SHM.

11.
Free Body Diagram Mass on

Spring

• Remember Hooke’s Law?

• F = -kx.

• F is the restoring force of the spring.

• Complete sheet.

Spring

• Remember Hooke’s Law?

• F = -kx.

• F is the restoring force of the spring.

• Complete sheet.

12.
Hwk Intro

13.
Simple Harmonic Motion

2 Conditions.

• 1. Acceleration/Fnet proportional to displacement.

• 2. Acceleration/Fnet directed toward equilibrium.

• Defining Equation for SHM

• a = -2x

2 Conditions.

• 1. Acceleration/Fnet proportional to displacement.

• 2. Acceleration/Fnet directed toward equilibrium.

• Defining Equation for SHM

• a = -2x

14.
Graphs Of SHM

15.
Acceleration - displacement

• Since F = - kx.

• ma = - kx.

• a = -k x

• m

• a –x

• Graph a (Y axis) vs. displacement on the X

• Since F = - kx.

• ma = - kx.

• a = -k x

• m

• a –x

• Graph a (Y axis) vs. displacement on the X

16.
a = − kx

• Negative gradient = when displacement is positive,

acceleration and restoring force are negative

• ∴a ∝−x

• experimental evidence shows k = ω2, where ω is the

angular velocity, which is be ω = 2Πf :

• a = − ω 2x

• Negative gradient = when displacement is positive,

acceleration and restoring force are negative

• ∴a ∝−x

• experimental evidence shows k = ω2, where ω is the

angular velocity, which is be ω = 2Πf :

• a = − ω 2x

17.
Sketching and v=0 v=v v= MAX

interpreting graphs of 0

SHM x

-2.0 0.0 2.0

EXAMPLE: The displacement x vs. time t for a system

undergoing SHM is shown here.

x-black

(+) ( -) (+) ( -) (+) t v-red

Sketch in red the velocity vs. time graph.

SOLUTION: At the extremes, v = 0.

At x = 0, v = vMAX. The slope determines sign of vMAX.

interpreting graphs of 0

SHM x

-2.0 0.0 2.0

EXAMPLE: The displacement x vs. time t for a system

undergoing SHM is shown here.

x-black

(+) ( -) (+) ( -) (+) t v-red

Sketch in red the velocity vs. time graph.

SOLUTION: At the extremes, v = 0.

At x = 0, v = vMAX. The slope determines sign of vMAX.

18.
Sketching and interpreting v=0 v = vMAXv =

graphs of simple harmonic 0

motion x

-2.0 0.0 2.0

EXAMPLE: The displacement x vs. time t for a system

undergoing SHM is shown here.

x-black

v-red

(different

t scale)

Sketch in blue the acceleration vs. time graph.

a-blue

SOLUTION: Since a -x, a is just a reflection of x.

(different

scale)

Note: x is a sine, v is a cosine, and a is a – sine wave.

graphs of simple harmonic 0

motion x

-2.0 0.0 2.0

EXAMPLE: The displacement x vs. time t for a system

undergoing SHM is shown here.

x-black

v-red

(different

t scale)

Sketch in blue the acceleration vs. time graph.

a-blue

SOLUTION: Since a -x, a is just a reflection of x.

(different

scale)

Note: x is a sine, v is a cosine, and a is a – sine wave.

19.
Plot of displacement, velocity, acceleration,

on same graph.

on same graph.

20.
Phase and Phase Difference

Two waves in Phase

Two waves in Phase

21.
Phase and Phase Difference

Phase shift =

Phase shift =

22.
4. What is the phase shift?

23.
Phase Difference

• The graphs all have the same period, but

• velocity leads displacement by ¼ of a period 90o.

• Acceleration leads velocity by ¼ of a period 90o.

• Displacement is 90o out of phase with velocity and 180o out of phase

with acceleration.

• When the phase difference is 0o or 360o, the systems are “in phase”.

• The graphs all have the same period, but

• velocity leads displacement by ¼ of a period 90o.

• Acceleration leads velocity by ¼ of a period 90o.

• Displacement is 90o out of phase with velocity and 180o out of phase

with acceleration.

• When the phase difference is 0o or 360o, the systems are “in phase”.

24.
-2.0 0.0 2.0 x

EXAMPLE: The kinetic energy

vs. displacement for a system

undergoing SHM is shown in

the graph. The system consists

of a 0.125-kg mass on a spring.

(a) Determine the maximum

velocity of the mass.

When the kinetic energy is maximum, the velocity

is also maximum. Thus 4.0 = (1/ 2)mvMAX2 so that

4.0 = (1/ 2)(.125)vMAX2 vMAX = 8.0 ms-1.

EXAMPLE: The kinetic energy

vs. displacement for a system

undergoing SHM is shown in

the graph. The system consists

of a 0.125-kg mass on a spring.

(a) Determine the maximum

velocity of the mass.

When the kinetic energy is maximum, the velocity

is also maximum. Thus 4.0 = (1/ 2)mvMAX2 so that

4.0 = (1/ 2)(.125)vMAX2 vMAX = 8.0 ms-1.

25.
-2.0 0.0 2.0 x

EXAMPLE: The kinetic energy ET

vs. displacement for a system

undergoing SHM is shown in

the graph. The system consists EK

of a 0.125-kg mass on a spring.

(b) Sketch EP and determine the EP

total energy of the system.

Since EK + EP = ET = CONST, and since EP = 0 when EK

= EK,MAX, it must be that ET = EK,MAX = 4.0 J.

Thus the EP graph will be the “inverted” EK graph.

EXAMPLE: The kinetic energy ET

vs. displacement for a system

undergoing SHM is shown in

the graph. The system consists EK

of a 0.125-kg mass on a spring.

(b) Sketch EP and determine the EP

total energy of the system.

Since EK + EP = ET = CONST, and since EP = 0 when EK

= EK,MAX, it must be that ET = EK,MAX = 4.0 J.

Thus the EP graph will be the “inverted” EK graph.

26.
-2.0 0.0 2.0 x

EXAMPLE: The kinetic energy

vs. displacement for a system

undergoing SHM is shown in

the graph. The system consists

of a 0.125-kg mass on a spring.

(c) Determine the spring

constant k of the spring.

SOLUTION: Recall EP = (1/2)kx2.

Note that EK = 0 at x = xMAX = 2.0 cm. Thus

EK + EP = ET = CONST ET = 0 + (1/ 2)kxMAX2 so that

4.0 = (1/ 2)k 0.0202 k = 20000 Nm-1.

EXAMPLE: The kinetic energy

vs. displacement for a system

undergoing SHM is shown in

the graph. The system consists

of a 0.125-kg mass on a spring.

(c) Determine the spring

constant k of the spring.

SOLUTION: Recall EP = (1/2)kx2.

Note that EK = 0 at x = xMAX = 2.0 cm. Thus

EK + EP = ET = CONST ET = 0 + (1/ 2)kxMAX2 so that

4.0 = (1/ 2)k 0.0202 k = 20000 Nm-1.

27.
-2.0 0.0 2.0 x

EXAMPLE: The kinetic energy

vs. displacement for a system

undergoing SHM is shown in

the graph. The system consists

of a 0.125-kg mass on a spring.

(d) Determine the acceleration

of the mass at x = 1.0 cm.

From Hooke’s law, F = -kx we get

F = -20000(0.01) = -200 N.

From F = ma we get -200 = 0.125a a = -1600 ms-2.

EXAMPLE: The kinetic energy

vs. displacement for a system

undergoing SHM is shown in

the graph. The system consists

of a 0.125-kg mass on a spring.

(d) Determine the acceleration

of the mass at x = 1.0 cm.

From Hooke’s law, F = -kx we get

F = -20000(0.01) = -200 N.

From F = ma we get -200 = 0.125a a = -1600 ms-2.

28.
x

-2.0 0.0 2.0

EXAMPLE: A 4.0-kg mass is

placed on a spring’s end and

displaced 2.0 m to the right.

The spring force F vs. its

displacement x from equilibrium

is shown in the graph.

(a) How do you know that the

mass is undergoing SHM?

In SHM, a -x. Since F = ma, then F -x also.

The graph shows that F -x. Thus we have SHM.

-2.0 0.0 2.0

EXAMPLE: A 4.0-kg mass is

placed on a spring’s end and

displaced 2.0 m to the right.

The spring force F vs. its

displacement x from equilibrium

is shown in the graph.

(a) How do you know that the

mass is undergoing SHM?

In SHM, a -x. Since F = ma, then F -x also.

The graph shows that F -x. Thus we have SHM.

29.
x

-2.0 0.0 2.0

EXAMPLE: A 4.0-kg mass is F = -5.0 N

placed on a spring’s end and x = 1.0 m

displaced 2.0 m to the right.

The spring force F vs. its

displacement x from equilibrium

is shown in the graph.

(b) Find the spring constant of

the spring.

SOLUTION: Use Hooke’s law: F = -kx.

Pick any F and any x. Use k = -F / x.

Thus k = -(-5.0 N) / 1.0 m = 5.0 Nm-1.

-2.0 0.0 2.0

EXAMPLE: A 4.0-kg mass is F = -5.0 N

placed on a spring’s end and x = 1.0 m

displaced 2.0 m to the right.

The spring force F vs. its

displacement x from equilibrium

is shown in the graph.

(b) Find the spring constant of

the spring.

SOLUTION: Use Hooke’s law: F = -kx.

Pick any F and any x. Use k = -F / x.

Thus k = -(-5.0 N) / 1.0 m = 5.0 Nm-1.

30.
x

-2.0 0.0 2.0

EXAMPLE: A 4.0-kg mass is

placed on a spring’s end and

displaced 2.0 m to the right.

The spring force F vs. its

displacement x from equilibrium

is shown in the graph.

(c) Find the total energy of the

SOLUTION: Use ET = (1/2)kxMAX2. Then

ET = (1/2)kxMAX2 = (1/2) 5.0 2.02 = 10. J.

-2.0 0.0 2.0

EXAMPLE: A 4.0-kg mass is

placed on a spring’s end and

displaced 2.0 m to the right.

The spring force F vs. its

displacement x from equilibrium

is shown in the graph.

(c) Find the total energy of the

SOLUTION: Use ET = (1/2)kxMAX2. Then

ET = (1/2)kxMAX2 = (1/2) 5.0 2.02 = 10. J.

31.
x

-2.0 0.0 2.0

EXAMPLE: A 4.0-kg mass is

placed on a spring’s end and

displaced 2.0 m to the right.

The spring force F vs. its

displacement x from equilibrium

is shown in the graph.

(d) Find the maximum speed of

the mass.

SOLUTION: Use ET = (1/2)mvMAX2.

10. = (1/2) 4.0 vMAX2

vMAX = 2.2 ms-1.

-2.0 0.0 2.0

EXAMPLE: A 4.0-kg mass is

placed on a spring’s end and

displaced 2.0 m to the right.

The spring force F vs. its

displacement x from equilibrium

is shown in the graph.

(d) Find the maximum speed of

the mass.

SOLUTION: Use ET = (1/2)mvMAX2.

10. = (1/2) 4.0 vMAX2

vMAX = 2.2 ms-1.

32.
x

-2.0 0.0 2.0

EXAMPLE: A 4.0-kg mass is

placed on a spring’s end and

displaced 2.0 m to the right.

The spring force F vs. its

displacement x from equilibrium

is shown in the graph.

(e) Find the speed of the mass

when its displacement is 1.0 m.

SOLUTION: Use ET = (1/2)mv 2 + (1/2)kx 2. Then

10. = (1/2)(4)v 2 + (1/2)(5)12

v = 1.9 ms-1.

-2.0 0.0 2.0

EXAMPLE: A 4.0-kg mass is

placed on a spring’s end and

displaced 2.0 m to the right.

The spring force F vs. its

displacement x from equilibrium

is shown in the graph.

(e) Find the speed of the mass

when its displacement is 1.0 m.

SOLUTION: Use ET = (1/2)mv 2 + (1/2)kx 2. Then

10. = (1/2)(4)v 2 + (1/2)(5)12

v = 1.9 ms-1.

33.

34.
Graphical Treatment

Equations of SHM

Equations of SHM

35.
Displacement, x, against time

x = xo cos t

start point at max ampl.

** Set Calculator in Radians.

x = xo cos t

start point at max ampl.

** Set Calculator in Radians.

36.
Displacement against time

x = xosint

• Speed = d/t = 2R/T but

• 2so

• v = r but r the displacement x so.

• v = x.

x = xosint

• Speed = d/t = 2R/T but

• 2so

• v = r but r the displacement x so.

• v = x.

37.
Velocity against time v = vocost

Starting where? Midpoint = max

velocity.

Starting where? Midpoint = max

velocity.

38.
Equations of Graphs

• x = xo cos t x = xo sin t

• v = -vo sin t v = vo cos t

• a = -aocos t -ao sin t

• Released from Released equilibrium.

top.

• x = xo cos t x = xo sin t

• v = -vo sin t v = vo cos t

• a = -aocos t -ao sin t

• Released from Released equilibrium.

top.

39.
Ex 2. A mass on a spring is oscillating

with f = 0.2 Hz and xo = 3 cm. What is

the displacement of the mass 10.66 s

after its release from the top?

• x = xo cos t xo = 3 cm

= 2f. = 0.4 Hz =1.26 rad/s.

• t = 10.66 s

• x = 0.03 cos (x 10.66) = 0.019 m

• You must use radians on calculator.

with f = 0.2 Hz and xo = 3 cm. What is

the displacement of the mass 10.66 s

after its release from the top?

• x = xo cos t xo = 3 cm

= 2f. = 0.4 Hz =1.26 rad/s.

• t = 10.66 s

• x = 0.03 cos (x 10.66) = 0.019 m

• You must use radians on calculator.

40.
SHM and Circular Motion

• Use the relationship to derive

equations.

• Use the relationship to derive

equations.

41.
If an object moving with constant speed in a

circular path is observed from a distant point (in

the plane of the motion), it will appear to be

oscillating with SHM.

The shadow of a pendulum bob moves with

s.h.m. when the pendulum itself is either

oscillating (through a small angle) or moving in

a circle with constant speed, as shown in the

circular path is observed from a distant point (in

the plane of the motion), it will appear to be

oscillating with SHM.

The shadow of a pendulum bob moves with

s.h.m. when the pendulum itself is either

oscillating (through a small angle) or moving in

a circle with constant speed, as shown in the

42.
For any s.h.m. we can find a corresponding circular

When a circular motion "corresponds to" a given s.h.m.,

i) the radius of the circle is equal to the amplitude

of the s.h.m.

ii) the time period of the circular motion is equal

to the time period of the s.h.m.

When a circular motion "corresponds to" a given s.h.m.,

i) the radius of the circle is equal to the amplitude

of the s.h.m.

ii) the time period of the circular motion is equal

to the time period of the s.h.m.

43.
Derive Relationship between accl & for SHM.

From circular motion ac = v2/r and vc = 2r/T

Oscillating systems have acceleration too.

• But = 2/T

• vc = 2r/T

• vc = r

• But ac = v2/r

• So ac = (r)2/r

• but r is related to displacement x.

From circular motion ac = v2/r and vc = 2r/T

Oscillating systems have acceleration too.

• But = 2/T

• vc = 2r/T

• vc = r

• But ac = v2/r

• So ac = (r)2/r

• but r is related to displacement x.

44.
For any displacement:

a = -²x

ao = -²xo

The negative sign shows Fnet & accl

direction opposite displacement.

Derivation of accl in Hamper pg 76.

Or use 2nd derivative of displacement.

a = -²x

ao = -²xo

The negative sign shows Fnet & accl

direction opposite displacement.

Derivation of accl in Hamper pg 76.

Or use 2nd derivative of displacement.

45.
Ex 3. A pendulum swings with f = 0.5 Hz. What

is the size & direction of the acceleration when

the bob has displacement of 2 cm right?

• a = -²x

• = 2f =

• a = -()2 (0.02 m) = -0.197 m/s2. left.

is the size & direction of the acceleration when

the bob has displacement of 2 cm right?

• a = -²x

• = 2f =

• a = -()2 (0.02 m) = -0.197 m/s2. left.

46.
Ex 4: A mass is bobbing on a spring with a

period of 0.20 seconds. What is its angular

acceleration at a point where its

displacement is 1.5 cm?

• = 2/T

• . = 31 rad/s

• a = -²x

• a = (31rad/s)(1.5 cm) = 1480 cm/s2.

• 15 m/s2.

period of 0.20 seconds. What is its angular

acceleration at a point where its

displacement is 1.5 cm?

• = 2/T

• . = 31 rad/s

• a = -²x

• a = (31rad/s)(1.5 cm) = 1480 cm/s2.

• 15 m/s2.

47.
To find the velocity of an oscillating mass or pendulum at any displacement:

When the mass is at equilibrium, x = 0, and

velocity is maximum:

vo = ± xo.

Derivation on H pg 77.

When the mass is at equilibrium, x = 0, and

velocity is maximum:

vo = ± xo.

Derivation on H pg 77.

48.
Ex 5. A pendulum swings with f = 1 Hz and amplitude 3 cm. At

what position will be its maximum velocity &what is the

At max velocity vo = xo.

f = 1) = rad/s

vo = (2 rad/s)(0.03)

vo = 0.188 m/s

vo = 0.2 m/s

what position will be its maximum velocity &what is the

At max velocity vo = xo.

f = 1) = rad/s

vo = (2 rad/s)(0.03)

vo = 0.188 m/s

vo = 0.2 m/s

49.
Hwk Hamper pg 75- 77 Show equations

and work, hand in virtual solar system lab.

Mechanical Universe w/questions

• http://www.learner.org/resources/series4

2.html?pop=yes&pid=565

and work, hand in virtual solar system lab.

Mechanical Universe w/questions

• http://www.learner.org/resources/series4

2.html?pop=yes&pid=565

50.
SHM, Hooke’s Law & k.

51.
For a mass undergoing SHM on a spring,

what is the relation between angular

frequency , and k the spring constant?

• Use Hooke’s law and make substitutions

to derive a relation in terms of angular

frequency, k, and mass.

what is the relation between angular

frequency , and k the spring constant?

• Use Hooke’s law and make substitutions

to derive a relation in terms of angular

frequency, k, and mass.

52.
F= - kx.

ma = - k x.

So a = -k x

m

a = -² x

So 2 = -k/m

ma = - k x.

So a = -k x

m

a = -² x

So 2 = -k/m

53.
How can you tell this graph depicts SHM?

Acceleration directly displacement, opposite direction.

What is the slope? 2 or k/m

Where is the amplitude?

Acceleration directly displacement, opposite direction.

What is the slope? 2 or k/m

Where is the amplitude?

54.
Energy in SHM

55.

56.
Pendulum is not SHM

• Fnet not directly opposite s

• for small displacement angles it approximates SHM.

• Fnet not directly opposite s

• for small displacement angles it approximates SHM.

57.
Energy Conservation in Oscillatory Motion

In an ideal system the total mechanical energy

is conserved. A horizontal mass on a spring:

Horizontal mass no PEg.

In an ideal system the total mechanical energy

is conserved. A horizontal mass on a spring:

Horizontal mass no PEg.

58.
Determining the max KE & PE:

• What is PE for a stretched spring.

• PE = ½ k x2 so PEmax =

• PE = ½ k xo

• What is the equation for KE?

• KE = ½ mv2

• Determine an equation for KEmax for SHM.

• vo = xo,

• KEmax = ½ m(2xo 2),

• What is PE for a stretched spring.

• PE = ½ k x2 so PEmax =

• PE = ½ k xo

• What is the equation for KE?

• KE = ½ mv2

• Determine an equation for KEmax for SHM.

• vo = xo,

• KEmax = ½ m(2xo 2),

59.
At any point:

• KE = ½ m2 (xo2 - x2 )

• How could you determine PE from Etot?

• Subtract KE from Etot.

• KE = ½ m2 (xo2 - x2 )

• How could you determine PE from Etot?

• Subtract KE from Etot.

60.
Since the total E will always equal the

max KE (or PE), we can calculate the

number of Joules of total E from the

KE equation:

ET = ½ m2xo2

max KE (or PE), we can calculate the

number of Joules of total E from the

KE equation:

ET = ½ m2xo2

61.
Ex 5: A 200-g pendulum bob is oscillating with

Amplitude = 3 cm, and f = 0.5 Hz. How much

KE will it have as it passes through the origin?

• KEmax = ½m 2xo 2,

• xo = 0.03 m

= .

• KE = 8.9 x 10-4J.

Amplitude = 3 cm, and f = 0.5 Hz. How much

KE will it have as it passes through the origin?

• KEmax = ½m 2xo 2,

• xo = 0.03 m

= .

• KE = 8.9 x 10-4J.

62.
6. The amplitude of an oscillating mass on

a vertical spring is 8.0 cm. The spring

constant is 74 N/m. Find the total energy of

the oscillator and the PE and KE at a

displacement = 4.8 cm.

• Et = 0.24 J.

• Ek = Et – Ep = 0.16 J.

a vertical spring is 8.0 cm. The spring

constant is 74 N/m. Find the total energy of

the oscillator and the PE and KE at a

displacement = 4.8 cm.

• Et = 0.24 J.

• Ek = Et – Ep = 0.16 J.

63.
Energy Conservation in Oscillatory Motion

The total energy is constant; as the kinetic

energy increases, the potential energy

decreases, and vice versa.

The total energy is constant; as the kinetic

energy increases, the potential energy

decreases, and vice versa.

64.
Energy Conservation in Oscillatory Motion

The E transforms from potential to kinetic &

back, the total energy remains the same.

The E transforms from potential to kinetic &

back, the total energy remains the same.

65.
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.

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.

66.
The Pendulum

The restoring force is

actually proportional

to sin θ, whereas the

restoring force for a

spring is proportional

to the displacement x.

The restoring force is

actually proportional

to sin θ, whereas the

restoring force for a

spring is proportional

to the displacement x.

67.

68.
Units

•The Pendulum

• Damped Oscillations

• Driven Oscillations and Resonance

•The Pendulum

• Damped Oscillations

• Driven Oscillations and Resonance

69.
Mass on a Spring

Since the force on a mass on a spring is

proportional to the displacement, and also to

the acceleration,

Make substitutions to find the relationship

between T and k.

Since the force on a mass on a spring is

proportional to the displacement, and also to

the acceleration,

Make substitutions to find the relationship

between T and k.

70.
Given that for a mass on a spring

Derive an equation that relates the period

of the oscillation to k, and m.

Derive an equation that relates the period

of the oscillation to k, and m.

71.
The Period of a Mass on a Spring

Therefore, the period is

How does T change as mass increases? Sketch it!

Therefore, the period is

How does T change as mass increases? Sketch it!

72.
Mass on Spring

73.
Period of pendulum

74.
The Pendulum

Substituting θ for sin θ allows us to treat the

pendulum in a mathematically identical way to

the mass on a spring.

Therefore, we find that the period of a pendulum

depends only on the length of the string:

Substituting θ for sin θ allows us to treat the

pendulum in a mathematically identical way to

the mass on a spring.

Therefore, we find that the period of a pendulum

depends only on the length of the string:

75.
The Pendulum

For small angles, sin θ and θ are approximately

For small angles, sin θ and θ are approximately

76.
Damped Oscillations

In most physical situations, there is a force,

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:

In most physical situations, there is a force,

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:

77.
Damped Oscillations

This exponential decrease is shown in the figure:

The image shows a system that is lightly damped – it goes

through multiple oscillations before coming to rest.

This exponential decrease is shown in the figure:

The image shows a system that is lightly damped – it goes

through multiple oscillations before coming to rest.

78.
Damped Oscillations

A critically damped system is one that

relaxes back to the equilibrium position

without oscillating through it. An

overdamped system will also not oscillate

but is damped so heavily that it takes

longer to reach equilibrium.

A critically damped system is one that

relaxes back to the equilibrium position

without oscillating through it. An

overdamped system will also not oscillate

but is damped so heavily that it takes

longer to reach equilibrium.

79.
Driven Oscillations and Resonance

An oscillation can be driven by an oscillating

driving force; the f of the driving force may or may

not be the same as the natural f of the system.

An oscillation can be driven by an oscillating

driving force; the f of the driving force may or may

not be the same as the natural f of the system.

80.
Driven Oscillations and Resonance

If the driving f is close

to the natural

frequency, the

amplitude can become

quite large, especially

if the damping is small.

This is called

If the driving f is close

to the natural

frequency, the

amplitude can become

quite large, especially

if the damping is small.

This is called

81.
Summary

• Period: time required for a motion to go

through a complete cycle

• Frequency: number of oscillations per unit time

• Angular frequency:

• Simple harmonic motion occurs when the

restoring force is proportional to the

displacement from equilibrium.

• Period: time required for a motion to go

through a complete cycle

• Frequency: number of oscillations per unit time

• Angular frequency:

• Simple harmonic motion occurs when the

restoring force is proportional to the

displacement from equilibrium.

82.
Summary

• The amplitude is the maximum displacement

from equilibrium.

• Position as a function of time:

• Velocity as a function of time:

• The amplitude is the maximum displacement

from equilibrium.

• Position as a function of time:

• Velocity as a function of time:

83.
Summary

• Acceleration as a function of time:

• Period of a mass on a spring:

• Total energy in simple harmonic motion:

• Acceleration as a function of time:

• Period of a mass on a spring:

• Total energy in simple harmonic motion:

84.
Summary

• Potential energy as a function of time:

• Kinetic energy as a function of time:

• A simple pendulum with small amplitude

exhibits simple harmonic motion

• Potential energy as a function of time:

• Kinetic energy as a function of time:

• A simple pendulum with small amplitude

exhibits simple harmonic motion

85.
Summary

• Period of a simple pendulum:

• Period of a physical pendulum:

• Period of a simple pendulum:

• Period of a physical pendulum:

86.
Summary

• An oscillating system may be driven by an

external force.

• This force may replace energy lost to friction,

or may cause the amplitude to increase greatly

at resonance

• Resonance occurs when the driving frequency

is equal to the natural frequency of the system

• An oscillating system may be driven by an

external force.

• This force may replace energy lost to friction,

or may cause the amplitude to increase greatly

at resonance

• Resonance occurs when the driving frequency

is equal to the natural frequency of the system

87.
Summary

• Oscillations where there is a nonconservative

force are called damped.

• Underdamped: the amplitude decreases

exponentially with time:

• Critically damped: no oscillations; system

relaxes back to equilibrium in minimum time

• Overdamped: also no oscillations, but

slower than critical damping

• Oscillations where there is a nonconservative

force are called damped.

• Underdamped: the amplitude decreases

exponentially with time:

• Critically damped: no oscillations; system

relaxes back to equilibrium in minimum time

• Overdamped: also no oscillations, but

slower than critical damping

88.

89.