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Waves

& Sound

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Objectives

• FCAT – Periodicity of waves– Movement of particles in transverse vs

longitudinal wave

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Sunshine State Standards

• SC.H.3.4.2 Students know that technological problems often create a demand for new scientific knowledge and that new technologies make it possible for scientists to extend their research in a way that advances science.

• SC.H.3.4.6 Students know that scientific knowledge is used by those who engage in design and technology to solve practical problems, taking human values and limitations into account

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Chapters 14 & 15

• Waves are periodic disturbances that propagate through a medium or space– a medium does not travel with the wave

• Mechanical waves require a medium

• Electromagnetic waves do not require a medium

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Key Formulae

Periodic MotionF = -kx PE = ½ kx2 T = 2 L/g

Waves f = 1/T v = f I = P/A = 10 log I fd = fs v + vd

Io v + vs

• V T inoC = 331.5 m/s + (0.6m/soC)(T)

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Assignments

P&P14 Waves

1-3,6,7,9,10,15-19,31,33,35,40,41,51,52,79

P&P15 Sound

1-3,6-8,12-14,

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Periodic Motion

• is repetitive motion• Simple harmonic motion is the result of a

restoring force on an object being directly proportional to the object’s displacement from equilibrium. That type of force obeys Hooke’s Law: F = -kx

• F, force, n k, spring constant, n/m• X, distance, m• Ignore the “–” (means restoring force

working against applied force)

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Sample Problem

Page 378.

1. How much force is necessary to stretch a spring 0.25 m when the spring constant is 95 N/m?

F = kx

F = (95 N/m)(0.25m)

F = 24 N

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Pendulums T, period, sec

L, length of pendulum, m

g, gravity, 9.8 m/s2 on Earth

What is the period on Earth of a pendulum with a length of 1.0 m?

= 2 sr(1.0m / 9.8 m/s2) = 2.0s

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Measuring Waves

, (lambda), wavelength, m

v, speed, m/s

f, frequency, 1/s, hertz

f = 1/T f, frequency, 1/s

T, period, s

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p. 386 15. A sound wave produced by a clock chime is heard 515 m away 1.50 s later.

a.What is the speed of sound of the clock’s chime in air?

V = d/t = 515 m / 1.50 s = 343 m/s

b.The sound wave has a frequency of 436 Hz. What is the period of the wave?

t = 1/f = 1/436 Hz = 2.29 x 10-3 s

c.What is the wave’s wavelength? = V / f = 343 m/2 / 436 Hz = 0.787 m

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More on waves:

• Longitudinal waves – particles of the medium move parallel to the direction of the wave

Sound is an example of this type of wave. Sound also requires a medium, categorizing it as a mechanical wave.

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Transverse Waves

• Transverse waves – displacement of the particles of the medium are perpendicular to the direction of propagation of the wave.

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Phase of a wave

physics.mtsu.edu

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Frequency

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Properties of Waves

• 1. Rectilinear propagation – advancement of a wave is perpendicular to the wave front

• 2. Reflection – waves bounce off barriers and rebound in opposite direction– Law of reflection:

• Incident angle = reflected angle (i = r)

• 3. Refraction – bending (changing direction) of a wave as it travels from one medium into another

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More properties…

• 4. Diffraction – spreading of a wave as it passes beyond the edge of a barrier

• http://www.physicsclassroom.com/Class/waves/U10L3b.html

• 5. Interference – result of 2 or more waves passing through the same medium at the same time

http://www.physicsclassroom.com/Class/waves/U10L3c.html

Refraction – bending after passing from one substance into another

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Source: chemicalparadigms.wikispaces.com

Source: matter.org.uk

Diffraction – spreading after passing by a barrier

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geographyfieldwork.com

Source: newgeology.us

Interference

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Source: discoverhover.org

Constructive – in phase, increase of amplitude, sounds get louder

Destructive – out of phase, cancel one another out

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Math associated with waves

• Frequency = 1 / Period (f = 1/T)

• Period = 1 / Frequency• i.e., f = 60cycles per second = 60 hz = 60/s

• Also:v = f

• where velocity = frequency x wavelength

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Sound Waves

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Math and waves

I = P/A Intensity = Power/Area

Units: watt / cm2

Intensity relates to loudness

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More math…

= 10 log I

Io 10-16

w/cm 2

Intensity level, decibels

I, Intensity , w/cm2

Io , threshhold of hearing, w/cm2

We all don’t hear the same, so this is a comparative measurement in decibels

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Intensity Problem

Sound energy is radiated uniformly in all directions from a small source at a rate of 1.2 watts.

A) What is the intensity of the sound at a point 25 m from the source?

Assume no energy is lost in transmission over a spherical area with radius of 25 m.

P = 1.2 w r = 25 m

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Finding the intensity, I

I = P = P

A 4r2

= 1.2 w

4 (2500 cm)2

= 1.5 x 10-8 w/cm2Source: hyperphysics.phy-astr.gsu.edu

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Find intensity level,

= 10 log I

Io

= 10 log 1.5 x 10-8 w/cm2

10-16 w/cm2

= 10 log (1.5x108 ) = 82 db

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Intensity Level to Intensity

Let’s say that the intensity level of a sound is 25.3 dB. What is the intensity of the sound in w/cm2?

= 10 log I / Io

25.3 dB = 10 log (I/10-16 w/cm2)

2.53 = log I – log 10-16

2.53 + log 10-16 = log I

2.53 – 16 = log I

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25.3 dB = 10 log (I/10-16 w/cm2)2.53 = log I – log 10-16

2.53 + log 10-16 = log I what power do you raise 10 to, to get 10-16?

2.53 – 16 = log I

Adding on the left -13.5 = log I

Raise 10 to the -13.5 power by this sequence:

2nd 10x (-13.5) 3.16 x 10 -14 w/cm2 = I

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More to come…Doppler Effect is the apparent change in frequency

as a result of relative motion between the source of a sound and a detector.

f d = f s v v d

v v s

fd, frequency heard by detector

fs, frequency of source

v, velocity of sound in air

vd, velocity of detector

vs, velocity of source

Source: onlinephys.com

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More on Doppler

Stationary detector:

Source moving away: v+vs on bottom

Source moving toward: v-vs on bottom

Stationary source:

detector moving toward: v+vd on top

detector moving away: v-vd on top

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Velocity of sound at various temps. *

• V T inoC = 331.5 m/s + (0.6m/soC)(T)

• 331.5 m/s is speed of sound in air at 0oC• 0.6m/soC rate of change of speed peroC change

What is the speed of sound in air at 10.00oC?

#1 V10oC = 331.5m/s +(0.6m/soC)(10oC)

V = 337.5 m/s

#2 V-5.2oC = 331.5 m/s + (0.6m/soC)(C)

V = 331.5 - 3.12 = 328 m/s

*Notice the 2 constants in red.

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Doppler Example

A stationary civil defense siren has a

frequency of 1000 Hz. What frequency

will be heard by drivers of cars moving

at 15 m/s?

A) away from the siren?

B) toward the siren?

The velocity of sound in air is 344 m/s.

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Solution: Moving away from the siren

fd = fs v - vd =

v

fd = (1000 Hz) (344 m/s - 15 m/s)

344 m/s

fd= 956 Hz

Apparent frequency heard by the detector decreases

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Solution: Moving toward the siren

fd = fs v + vd = v

f d = (1000 Hz) (344 m/s + 15 m/s) 344 m/s

f d = 1044 Hz

Apparent frequency increases

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One More…

A police car with a 1000 Hz siren is

moving at 15 m/s. What frequency

is heard by a stationary listener when

the police car is

a) receding from the detector?

b) approaching the detector?

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f d = f s v

v + v s

f d = (1000 Hz) 344 m/s = 958 Hz

344 m/s + 15 m/s P. car moving away

f d = f s v

v - v s

f d = (1000 Hz) 344 m/s = 1046 Hz

344 m/s - 15 m/s P. car coming toward