Waves

andSound

Simple Harmonic Motion

period, T:

frequency, f:

time to complete one cycle

# of cycles per second (Hz)

f1 T

T1

f

(s)

wall equilibrium position

frictionless floor

x = 0 Felas = 0

xmax

–xmax

v = 0 x = max

Felas = max

period of a mass-spring system:

km

2 T m = mass (kg) k = spring constant (N/m)

A 5.5 kg cat is attached to a fixedhorizontal spring of stiffness 22.8 N/mand is set in motion on a frictionlesssurface. Find the period of motion of…

…the cat.

…a 240 g mouse, with the same spring and surface.

km

2 T m

N 22.8

kg 5.52 = 3.1 s

kg 0.24 g 10 x 1

kg 1 g 240 m 3

km

2 T m

N 22.8

kg 0.242 = 0.64 s

= 0.99 N/m

What stiffness must a spring have so that the periodof the mouse’s motion is the same as that of the cat?

km

2T

km

2T k

m4T

2

2

m4 kT 22

2

2

Tm4

k

22

1.3

24.04

s

kg

Ballpark answer: Need a “less stiff” spring; k < 22.8 N/m.

= 1.2 Hz

A 1275 kg car carries two passengerswith a combined mass of 153 kg. Thecar has four shock absorbers, each witha spring constant of 2.0 x 104 N/m. Findthe frequency of the vehicle’s motionafter it hits a pothole.

N/m) 10 x 4(2.0kg)1531275(

2

1

4

km

2 T

km2

1 f

restoring force:

simple harmonic motion (SHM):

For a mass-spring system,Hooke’s law applies:

acts to move an objectback to equilibrium

Frestore x

Frestore = Felas = k x

As displacement increases, so does Frestore.

And when x = 0… Frestore = 0.

KE = ½ m v2

energy:

kinetic energy, KE: energy of mass m having velocity v

m (kg); v (m/s)

the ability to do work

KE = max. at eq. pos.; KE = 0 at xmax.

PEelas = ½ k (x)2

potential energy, PE: stored energy

For a spring with spring constant k and “stretch” x:

k (N/m); x (m)

For a mass m at a height h above a reference line:

m (kg); h (m);g = 9.81 m/s2

For m-s sys., PEelas is max at xmax and 0 at eq. pos.

PEg = m g h

For pend., PEg is max atxmax and min. at eq. pos.

amplitude, A: maximum displacement from equilibrium

Energy of a Mass-Spring System

frictionless

A A

Period T is not affected by amplitude A.

Energy (J)

total energy

PEelas

KE PEg

km

2T

The Pendulum

For < 15o, asimple pendulum

approximates SHM. mass mof bob

length Lamplitude

Energy of a Simple Pendulum

Energy (J)

total energy

PEg

KE PEelas

period of a simple pendulum:

The period of a pendulum is 5.2 s. Find…

A. …its length

B. …the mass of the bob

g L

2 T

Period T is independent of mass and amplitude.

g L

2 T g L

2T

gL

4T

2

2

2

2

4gT

L

2

22

4

s) 2.5( s

m 9.81

= 6.7 m

NOT ENOUGH INFORMATION

Waves

vibrations moving through space and time

medium: the matter through which the energy of mechanical waves moves

Waves transmit energy, not matter.

transverse waves: particles of medium move to direction of wave travel

For a transverse wave:

amplitude A crest

wavelength

trough

Energy Amplitude2

longitudinal(compressional) wave:

particles of mediummove // to directionof wave travel

rarefaction compression

pulse wave:

periodic wave:

a single vibration

rhythmic, repeated vibrations

Wave Reflection

fixed boundary free boundary

v

v

v

v

A A

waves are reflectedand inverted

waves are reflectedand upright

Wave Interference

Two waves (unlike two objects) can occupythe same place at the same time. This

condition is called interference.

constructive interference: destructive interference: displacements arein same direction

displacements arein opposite directions

A1 A2

A1 + A2

A1 A2

A2

A1

A1 – A2

A2

A1

= 1.8 m/s

Wave Velocity

Equation: v (m/s); (m);T (s); f (Hz)

f

T v

A water wave of wavelength 8.5 mwashes past a boat at anchor every4.75 seconds. Find the wave’s velocity.

T

v

s 4.75m 8.5

vsound = 331 + 0.6Ta

The velocity of any mechanical wavedepends only on properties of medium

through which it travels.

e.g., string tension,water depth,air temperature,material density,type of material

An empirical equation for thevelocity of sound in air:

Ta = air temp. in oC

Standing Waves

incident and reflected waves interfere so thatantinodes have a max. amplitude, while nodeshave zero amplitude

On a string, nodes remain motionless; antinodesgo from max. (+) to max. (–) displacement.

wavelength of nth

harmonic on a string:

n = 1; 1st harmonic (fundamental)

n = 2; 2nd harmonic(1st overtone)

n = 3; 3rd harmonic(2nd overtone)

n = 4; 4th harmonic(3rd overtone)

nL 2

n (n = 1,2,3,…)

1 = 2 L

2 = L

3 = 2/3 L

4 = ½ L

L

fn = n f1

Waves travel along a 96.1 cmguitar string at 492 m/s. Find thefundamental frequency of the string.

Find the frequency ofthe 5th harmonic.

frequency of the nth harmonic:

1 = 2 L = 2 (0.961 m) = 1.922 m

v = f and v = f1 1

11

v f

m 922.1

m/s 492 = 256 Hz

5 = 2/5 L = 0.3844 m

55

v f

m .38440

m/s 492 = 1280 Hz

nL 2

n

Standing Waves in Open Tubes

wavelength of nth

harmonic of an open tube:

n = 1

L

1 = 2 L n = 2 2 = L

n = 3 3 = 2/3 L

nL 2

n (n = 1,2,3,…)

Closed Tubes

wavelength of nth

harmonic of a closed tube:

n = 1

L

1 = 4 L n = 2

2 = 2 Ln = 3

3 = 4/3 L

nL 4

n (n = 1,3,5,…)

(even harmonicsare not present)

Find the fundamental frequencyfor an open tube of length 1.24 m.Assume the air temperature tobe 20.0oC.

1.24 m

nL 2

n

1 = 2 L = 2 (1.24 m) = 2.48 m

v = f and v = f1 1

11

v f

m 48.2m/s 343

f1 = 138 Hz

v = ?vsound = 331 + 0.6Ta

vsound = 331 + 0.6(20) = 343 m/s

Find fundamental frequency fora closed tube of length 1.24 m.Air temp. is 20.0oC.

nL 4

n

1 = 4 L = 4 (1.24 m) = 4.96 m

v = f and v = f1 1

11

v f

m 96.4

m/s 343 = 69.0 Hz

1.24 m

(same as prev. prob.)

doubling (or halving)of frequency

=

100 Hz 200 Hz 400 Hz 800 Hz (three octaves)

one octave

Sound

compression:

rarefaction:

20 Hz 20,000 Hz

high pressure / high density

low pressure / low density

audiblefrequencies

(humanhearing)infrasonic ultrasonic

Fundamental frequency determines pitch.

high pitch high f =

low f =

= short

low pitch = long

Number and intensity of an instrument’s harmonicsgive it its unique sound quality, or ________.timbre

f1 f2 f3 f4 f1 f2 f3 f4

f1 f2 f3 f4 f1 f2 f3 f4

The Doppler Effect Relative motion between wave source and observercauses a change in the ____________ frequency. observed

fobserved

fobserved

(higher)

(lower)

femitted femitted

femitted

v = 0

Other examples of Doppler effect:

police radar

race cars

expansion of universe dolphins (echolocation)

R O Y G B V

Sun

most stars (“red-shifted”)

Traveling Very Fast

vbug = 0 vbug < vwave

vbug = vwave vbug > vwave

bow wavewave barrier

supersonic: “faster than sound” (vs. subsonic)

sonic boom:

shock wave: a 3-D bow wave

caused by high-pressureair, not roaring engine

cracking bulletslion tamer’s whip

The Matrix

= 0.024 W/m2

Sound Intensity

If a piano’s power output is 0.302 W,find the sound intensity at a distance of…

A. …1.0 m

B. …2.0 m

r 4

P I 2

2r 4P

I

2m) (1.0 4 W0.302

= 0.0060 W/m22r 4

P I

2m) (2.0 4

W0.302

Intensity is related to volume (or relative intensity):

--

-- measured in decibels (dB)

A difference of 10 dB changesthe sound intensity

by a factor of 10 andthe volume by a factor of 2.

50 dB 40 dB 60 dB 90 dB

how loud we perceive a sound to be

half as loud 1/10 as intense

8X louder

1000X more intense

Beats

alternating loud-and-soft sounds resultingfrom interference between two slightly-different frequencies

Equation: f - f f 21beat

1 second elapses 1 second elapses

f 1 =

16

Hz

f 2 =

18

Hzf 1 =

16

Hz

f 2 =

17

Hz

fbeat = 1 Hz fbeat = 2 Hz

Forced Vibrations and Resonance

natural frequency:

forced vibration:

resonance:

-- result of resonance =

the frequency at which anobject most easily vibrates

a vibration due toan applied force

occurs when a force is repeatedlyapplied to an object AT the object’s natural frequency

large amplitude

Examples:

swing

shattering crystal wine glasses

Tacoma Narrows Bridge (1940)

British regiment (Manchester, 1831)

aeolian harps

“The wind in the wires made a tattletale sound,as a wave broke over the railing…”

f1 T

T1

f

km

2 T

Frestore = Felas = k x

KE = ½ m v2

PEelas = ½ k (x)2

PEg = m g h

g L

2 T

f

T v

vsound = 331 + 0.6Ta

nL 2

n

fn = n f1

nL 4

n

r 4

P I 2

f - f f 21beat

h

h