02 Waves Review.pptx

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Mechanical Waves

A mechanical wave is a physical disturbance in an elastic medium.

Consider a stone dropped into a lake.

Energy is transferred from stone to floating log, but only the disturbance travels.

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Periodic Motion

Simple periodic motion is that motion in which a body moves back and forth over a fixed path, returning to each position and velocity after a definite interval of time.

Amplitude A

Period, T, is the time for one complete

oscillation. (seconds,s)

Period, T, is the time for one complete

oscillation. (seconds,s)

Frequency, f, is the number of complete oscillations per

second. Hertz (s-1) Frequency, f, is the number of complete oscillations per

second. Hertz (s-1)

1

f

T

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Review of Simple

Harmonic Motion

x F

It might be helpful for

you to review Chapter 14 on Simple Harmonic

Motion. Many of the same terms are used in this

chapter.

1

2

k

f

m





T

2

m

k



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Example: The suspended mass makes 30 complete

oscillations in 15 s. What is the period and frequency of the motion?

x F

Period: T = 0.500 s

Frequency: f = 2.00 Hz

15 s

0.50 s

30 cylces

T



1

0.500 s

f

T

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A Transverse Wave

In a transverse wave, the vibration of the individual particles of the medium is

perpendicular to the direction of wave propagation.

In a transverse wave, the vibration of the individual particles of the medium is

perpendicular to the direction of wave propagation.

Motion of particles

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Longitudinal Waves

In a longitudinal wave, the vibration of the individual particles is parallel to the

direction of wave propagation.

In a longitudinal wave, the vibration of the individual particles is parallel to the

direction of wave propagation.

Motion of particles

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Water Waves

An ocean wave is a combi-nation of transverse and longitudinal.

The individual particles move in ellipses as the wave disturbance moves toward the shore.

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Wave speed in a string.

v = speed of the transverse wave (m/s)

F = tension on the string (N)

m or m/L = mass per unit length (kg/m)

v = speed of the transverse wave (m/s)

F = tension on the string (N)

m or m/L = mass per unit length (kg/m)

The wave speed v in a vibrating string is determined by the tension F and the linear density m, or mass per unit length.

L

m = m/L

F FL v

m

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Example 1: A 5-g section of string has a length of 2 m from

the wall to the top of a pulley. A 200-g mass hangs at the end. What is the speed of a wave in this string?

200 g F = (0.20 kg)(9.8 m/s2_{) = 1.96 N}

v = 28.0 m/s v = 28.0 m/s

Note: Be careful to use consistent units. The tension F must be in newtons, the mass m in kilograms, and the length L in meters.

Note: Be careful to use consistent units. The tension F must be in newtons, the mass m in

kilograms, and the length L in meters.

(1.96 N)(2 m) 0.005 kg

FL v

m

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Periodic Wave Motion

l

B A

Wavelength l is distance between two particles that are in phase.

A vibrating metal plate produces a

transverse continuous wave as shown.

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Velocity and Wave Frequency.

The period T is the time to move a distance of one wavelength. Therefore, the wave speed is:

The frequency f is in s-1 or hertz (Hz).

The velocity of any wave is the product of the frequency and the wavelength:

1

but so

v T v f

T f

l _l

  

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Production of a Longitudinal Wave

• An oscillating pendulum produces condensations

and rarefactions that travel down the spring.

• The wave length λ is the distance between

adjacent condensations or rarefactions. l

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Velocity, Wavelength, Speed

Frequency f = waves per second (Hz)

Velocity v (m/s)

Wavelength l (m) l Wave equation s v t 

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Example 2: An electromagnetic vibrator sends waves down a string. The vibrator makes 600 complete cycles in 5 s. For one complete vibration, the wave moves a distance of 20 cm.

What are the frequency, wavelength, and velocity of the wave?

f = 120 Hz f = 120 Hz

The distance moved during a time of one cycle is the

wavelength; therefore:

l = 0.20 m

v = fl

v = (120 Hz)(0.02 m)

v = 2.40 m/s v = 2.40 m/s

600 cycles ; 5 s

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Energy of a Periodic Wave

The energy of a periodic wave in a string is a

function of the linear density m , the frequency f, the velocity v, and the amplitude A of the wave.

f A

v

m = m/L

2 2 2

2

E

f A

L





m

2 2 2

2

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Example 3. A 2-m string has a mass of 300 g and

vibrates with a frequency of 20 Hz and an amplitude of

50 mm. If the tension in the rope is 48 N, how much power must be delivered to the string?

P = 22(20 Hz)2(0.05 m)2(0.15 kg/m)(17.9 m/s)

P = 53.0 W

0.30 kg

0.150 kg/m 2 m

m L

m   

(48 N) 17.9 m/s 0.15 kg/m F v m

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The Superposition Principle

• When two or more waves (blue and green) exist in

the same medium, each wave moves as though the other were absent.

• The resultant displacement of these waves at any

point is the algebraic sum (yellow) wave of the two displacements.

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Reflection of Waves

(FREE End Reflection)

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Reflection of Waves

(FIXED End Reflection)

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If you continue to make

waves, the

returning

waves will interfere with

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If you generate waves of the

“CORRECT”

wavelength,

the returning waves will

come back

“in step”

with

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The returning wave will

arrive at a

crest

the

moment you are making

the

crest

of the next wave.

“In Step” for

FREE

End

Constructive Interference will occur,

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Close your eyes and make a standing wave.

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The returning wave will

arrive at a

trough

the

moment you are making

the

crest

of the next wave.

Destructive Interference will occur, and

the end of the medium will

not

move.

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Possible Standing Waves

on a Fixed String

12 ft

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on a Fixed String

12 ft

Possible λ’s for a 12 ft medium

24

ft

, 12

ft

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on a Fixed String

12 ft

24

ft

, 12

ft

, 8

ft

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Possible Standing Waves

on a Fixed String

12 ft

24

ft

, 12

ft

, 8

ft

, 6

ft

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on a Fixed String

12 ft

24

ft

, 12

ft

, 8

ft

, 6

ft

, 4.8

ft

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on a Fixed String

24

ft

, 12

ft

, 8

ft

, 6

ft

, 4.8

ft

12 ft

Find the next TWO possible wavelengths!

Write to Learn

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on a Fixed String

12 ft

Possible λ’s for any length, L

24

ft

, 12

ft

, 8

ft

, 6

ft

, 4.8

ft

2.4 ft 2.4 ft 2.4 ft 2.4 ft 2.4 ft

2 · L

n

=

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Because each of these waves are

POSSIBLE,

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POSSIBLE,

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on a Fixed String

Fundamental Frequency

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on a Fixed String

(2

nd

Harmonic)

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on a Fixed String

(3

rd

Harmonic)

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on a Fixed String

(4

th

Harmonic)

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on a Fixed String

(5

th

Harmonic)

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POSSIBLE,

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FREE

End

FIXED

End

¼ λ

12 ft

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FREE

End

FIXED

End

¾ λ

4 ft 4 ft 4 ft

48

ft

, 16

ft

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FREE

End

FIXED

End

5

/

4

λ

2.4 ft 2.4 ft 2.4 ft 2.4 ft 2.4 ft

48

ft

, 16

ft

, 9.6

ft

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FREE

End

FIXED

End

7

/

4

λ

1.71 ft 1.71 ft 1.71 ft 1.71 ft 1.71 ft 1.71 ft 1.71 ft

48

ft

, 16

ft

, 9.6

ft

, 6.86

ft

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FREE

End

FIXED

End

9

/

4

λ

1.33 ft 1.33 ft 1.33 ft 1.33 ft 1.33 ft 1.33 ft 1.33 ft 1.33 ft 1.33 ft

12 ft

48

ft

, 16

ft

, 9.6

ft

, 6.86

ft

,

5.33

ft

Find the next TWO possible wavelengths!

Write to Learn

Question #4

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FREE

End

FIXED

End

Possible λ’s for any length, L

=

4 · L

2n-1

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Formation of a

Standing Wave:

Incident and reflected waves traveling in

opposite directions produce nodes N and antinodes A.

The distance between

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Possible Wavelengths for Standing Waves

Fundamental, n = 1

1st overtone, n = 2

2nd overtone, n = 3

3rd overtone, n = 4

n = harmonics

2

1, 2, 3, . . .

n

L

n n

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Characteristic Frequencies

Now, for a string under tension, we have:

Characteristic

frequencies: n 2 ; 1, 2, 3, . . .

n F f n L m   and 2

F FL nv

v f

m L

m

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Example 4. A 9-g steel wire is 2 m long and is under a tension of 400 N. If the string vibrates in three loops, what is the

frequency of the wave?

400 N For three loops: n = 3

f₃ = 224 Hz

Third harmonic 2nd overtone

; 3 2 n n F f n L m   3

3 3 (400 N)(2 m) 2 2(2 m) 0.009 kg

FL f

L m