chapter 13 vibrations and waves
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Chapter 13 Vibrations and Waves. Heinrich Hertz (1857-1894). Periodic motion Periodic ( harmonic ) motion – self-repeating motion Oscillation – periodic motion in certain direction Period (T) – a time duration of one oscillation - PowerPoint PPT PresentationTRANSCRIPT
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Chapter 13
Vibrations and Waves
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Periodic motion
• Periodic (harmonic) motion – self-repeating motion
• Oscillation – periodic motion in certain direction
• Period (T) – a time duration of one oscillation
• Frequency (f) – the number of oscillations per unit time, SI unit of frequency 1/s = Hz (Hertz)
Tf 1
Heinrich Hertz(1857-1894)
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Motion of the spring-mass system
• Hooke’s law:
• The force always acts toward the equilibrium position: restoring force
• The mass is initially pulled to a distance A and released from rest
• As the object moves toward the equilibrium position, F and a decrease, but v increases
kxF
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Motion of the spring-mass system
• At x = 0, F and a are zero, but v is a maximum
• The object’s momentum causes it to overshoot the equilibrium position
• The force and acceleration start to increase in the opposite direction and velocity decreases
• The motion momentarily comes to a stop at x = - A
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Motion of the spring-mass system
• It then accelerates back toward the equilibrium position
• The motion continues indefinitely
• The motion of a spring mass system is an example of simple harmonic motion
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Simple harmonic motion
• Simple harmonic motion – motion that repeats itself and the displacement is a sinusoidal function of time
)cos()( tAtx
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Amplitude
• Amplitude – the magnitude of the maximum displacement (in either direction)
)cos()( tAtx
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Phase
)cos()( tAtx
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Phase constant
)cos()( tAtx
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Angular frequency
)cos()( tAtx
)(coscos TtAtA 0
)2cos(cos )(cos)2cos( Ttt
T 2
T 2
f 2
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Period
)cos()( tAtx
2
T
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Velocity of simple harmonic motion
)cos()( tAtx
)sin()( tAtv
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Acceleration of simple harmonic motion
)cos()( tAtx
)cos()( 2 tAta
)()( 2 txta
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The force law for simple harmonic motion
• From the Newton’s Second Law:
• For simple harmonic motion, the force is proportional to the displacement
• Hooke’s law:
maF
kxF
xm 2
mk
kmT 22mk
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Energy in simple harmonic motion
• Potential energy of a spring:
• Kinetic energy of a mass:
2/)( 2kxtU )(cos)2/( 22 tkA
2/)( 2mvtK )(sin)2/( 222 tAm
)(sin)2/( 22 tkA km 2
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Energy in simple harmonic motion
)(sin)2/()(cos)2/( 2222 tkAtkA
)()( tKtU
)(sin)(cos)2/( 222 ttkA)2/( 2kA )2/( 2kAKUE
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Energy in simple harmonic motion
)2/( 2kAKUE
2/2/2/ 222 mvkxkA kmvxA /222
22 xAmkv 22 xA
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Chapter 13Problem 11
A simple harmonic oscillator has a total energy E. (a) Determine the kinetic and potential energies when the displacement is one-half the amplitude. (b) For what value of the displacement does the kinetic energy equal the potential energy?
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Pendulums
• Simple pendulum:
• Restoring torque:
• From the Newton’s Second Law:
• For small angles
)sin( gFL
I
sin
I
mgL
)sin( gFL
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Pendulums
• Simple pendulum:
• On the other hand
Lat
I
mgL
Ls
sI
mgLa
)()( 2 txta
ImgL
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Pendulums
• Simple pendulum:
ImgL
2mLI
2mLmgL
Lg
gLT
22
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Pendulums
• Physical pendulum:
Imgh
mghIT
22
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Chapter 13Problem 32
An aluminum clock pendulum having a period of 1.00 s keeps perfect time at 20.0°C. (a) When placed in a room at a temperature of –5.0°C, will it gain time or lose time? (b) How much time will it gain or lose every hour?
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Simple harmonic motion and uniform circular motion
• Simple harmonic motion is the projection of uniform circular motion on the diameter of the circle in which the circular motion occurs
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Simple harmonic motion and uniform circular motion
• Simple harmonic motion is the projection of uniform circular motion on the diameter of the circle in which the circular motion occurs
)cos()( tAtx
)sin()( tAtvx
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Simple harmonic motion and uniform circular motion
• Simple harmonic motion is the projection of uniform circular motion on the diameter of the circle in which the circular motion occurs
)sin()( tAtvx
)cos()( tAtx
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Simple harmonic motion and uniform circular motion
• Simple harmonic motion is the projection of uniform circular motion on the diameter of the circle in which the circular motion occurs
)cos()( tAtx
)cos()( 2 tAtax
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Damped simple harmonic motion
bvFb Dampingconstant
Dampingforce
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Forced oscillations and resonance
• Swinging without outside help – free oscillations
• Swinging with outside help – forced oscillations
• If ωd is a frequency of a driving force, then forced oscillations can be described by:
• Resonance:
)cos(),/()( tbAtx dd
d
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Forced oscillations and resonance
• Tacoma Narrows Bridge disaster (1940)
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Wave motion
• A wave is the motion of a disturbance
• All waves carry energy and momentum
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Types of waves
• Mechanical – governed by Newton’s laws and exist in a material medium (water, air, rock, ect.)
• Electromagnetic – governed by electricity and magnetism equations, may exist without any medium
• Matter – governed by quantum mechanical equations
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Types of waves
Depending on the direction of the displacement relative to the direction of propagation, we can define wave motion as:
• Transverse – if the direction of displacement is perpendicular to the direction of propagation
• Longitudinal – if the direction of displacement is parallel to the direction of propagation
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Types of waves
Depending on the direction of the displacement relative to the direction of propagation, we can define wave motion as:
• Transverse – if the direction of displacement is perpendicular to the direction of propagation
• Longitudinal – if the direction of displacement is parallel to the direction of propagation
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Superposition of waves
• Superposition principle – overlapping waves algebraically add to produce a resultant (net) wave
• Overlapping solutions of the linear wave equation do not in any way alter the travel of each other
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Sinusoidal waves
• One of the most characteristic solutions of the linear wave equation is a sinusoidal wave:
• A – amplitude, φ – phase constant
)2/)(cos())(sin()(
vtxkAvtxkAvtxy
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Wavelength
• “Freezing” the solution at t = 0 we obtain a sinusoidal function of x:
• Wavelength λ – smallest distance (parallel to the direction of wave’s travel) between repetitions of the wave shape
))(cos(),( vtxkAtxy
)cos()0,( kxAxy
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Wave number
• On the other hand:
• Angular wave number: k = 2π / λ
)cos()0,( kxAxy ))(cos( xkA
)cos( kkxA
)2cos()cos( kxkx /2k
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Angular frequency
• Considering motion of the point at x = 0 we observe a simple harmonic motion (oscillation) :
• For simple harmonic motion:
• Angular frequency ω
))(cos(),( vtxkAtxy
)cos(),0( kvtAty )cos( kvtA
)cos()( tAty
/2 vkv
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Frequency, period
• Definitions of frequency and period are the same as for the case of rotational motion or simple harmonic motion:
• Therefore, for the wave velocity
2//1 Tf /2T
fTkv //
)cos(),( tkxAtxy
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Wave velocity
• v is a constant and is determined by the properties of the medium
• E.g., for a stretched string with linear density μ = m/l under tension T
Tv
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Chapter 13Problem 41
A harmonic wave is traveling along a rope. It is observed that the oscillator that generates the wave completes 40.0 vibrations in 30.0 s. Also, a given maximum travels 425 cm along the rope in 10.0 s. What is the wavelength?
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Interference of waves
• Interference – a phenomenon of combining waves, which follows from the superposition principle
• Considering two sinusoidal waves of the same amplitude, wavelength, and direction of propagation
• The resultant wave:
)cos(),(2 tkxAtxy)cos(),(1 tkxAtxy
),(),(),( 21 txytxytxy
)cos()cos( tkxAtkxA
2
cos2
cos2coscos
)2/cos()2/cos(2 tkxA
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Interference of waves
• If φ = 0 (Fully constructive)
• If φ = π (Fully destructive)
• If φ = 2π/3 (Intermediate)
)2/cos()2/cos(2),( tkxAtxy
)cos(2),( tkxAtxy
0),( txy
)3/cos()3/cos(2),(
tkxAtxy
)3/cos( tkxA
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Reflection of waves at boundaries
• Within media with boundaries, solutions to the wave equation should satisfy boundary conditions. As a results, waves may be reflected from boundaries
• Hard reflection – a fixed zero value of deformation at the boundary – a reflected wave is inverted
• Soft reflection – a free value of deformation at the boundary – a reflected wave is not inverted
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Questions?
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Answers to the even-numbered problems
Chapter 13
Problem 2
(a) 1.1 × 102 N(b) The graph is a straight line passing
through the origin with slope equal to k = 1.0 × 103 N/m.
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Answers to the even-numbered problems
Chapter 13
Problem 8
(a) 575 N/m(b) 46.0 J
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Answers to the even-numbered problems
Chapter 13
Problem 12
(a) 2.61 m/s(b) 2.38 m/s
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Answers to the even-numbered problems
Chapter 13
Problem 16
(a) 0.15 J(b) 0.78 m/s(c) 18 m/s2