1 Waves & Sound 2 Objectives FCAT –Periodicity of waves –Movement of particles in transverse vs...
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Transcript of 1 Waves & Sound 2 Objectives FCAT –Periodicity of waves –Movement of particles in transverse vs...
1
Waves
& Sound
2
Objectives
• FCAT – Periodicity of waves– Movement of particles in transverse vs
longitudinal wave
3
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