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Chapter 13
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Hooke’s Law: F = - kx
Periodic & Simple Harmonic Motion
Springs & Pendula
Waves
Next Week!
Superposition
Review Physics 2A:
Springs, Pendula & Circular Motion
Elastic Systems F = −kx
Small Vibrations
Natural Frequency of Objects
& Resonance
All objects have a natural frequency of
vibration or oscillation. Bells, tuning
forks, bridges, swings and atoms all
have a natural frequency that is related
to their size, shape and composition.
A system being driven at its natural
frequency will resonate and produce
maximum amplitude and energy.
Coupled Oscillators
Molecules, atoms and particles are modeled as coupled oscillators.
Waves Transmit Energy through coupled oscillators.
Forces are transmitted between the oscillators like springs
Coupled oscillators make the medium.
The Oscillation of Nothing
Hooke’s Law
Ut tensio, sic vis - as the extension, so is the force
Hooke’s Law describes the elastic response to an applied force.
Elasticity is the property of an object or material which causes
it to be restored to its original shape after distortion.
An elastic system displaced from equilibrium oscillates in a
simple way about its equilibrium position with
Simple Harmonic Motion.
Robert Hooke (1635-1703)
•Leading figure in Scientific Revolution
•Contemporary and arch enemy of Newton
•Hooke’s Law of elasticity
•Worked in Physics, Biology, Meteorology, Paleontology
•Devised compound microscope
•Coined the term “cell”
Hooke’s Law
It takes twice as much force to stretch a spring twice as far.
The linear dependence of displacement upon stretching force:
Fapplied = kx
Hooke’s Law
Stress is directly proportional to strain.
Fapplied ( stress ) = kx( strain)
Hooke’s Law
FRestoring = − kx
•The applied force
displaces the system a
distance x.
•The reaction force of the
spring is called the
“Restoring Force” and it is
in the opposite direction to
the displacement.
An Ideal Spring
Fspring = − kx
An ideal spring obeys
Hooke’s Law and exhibits
Simple Harmonic Motion.
Spring Constant k: Stiffness
•The larger k, the stiffer the spring
•Shorter springs are stiffer springs
•k strength is inversely proportional to the number of coils
Spring Question
Each spring is identical with the same spring constant, k.
Each box is displaced by the same amount and released.
Which box, if either, experiences the greater net force?
Hooke’s Law: F = - k x
+
Hooke’s Law: F = - k x
+
Hooke’s Law: F = - k x
+
Hooke’s Law: F = - k x
+
Hooke’s Law: F = - k x
+
Periodic Motion
Position vs Time: Sinusoidal Motion
SHM and Circular Motion
x ( t ) = A cos θ ( t )
v(t ) = − Aω sin ω t
a (t ) = − Aω 2 cos ω t
2π R
v2
vt = Rω =
, at = Rα , ac =
= ω2R
R
Τ
http://www.edumedia-sciences.com/en/a270-uniform-circular-motion
Terms:
Amplitude : [ A] = m
Period : T = time / cycle, [T ] = sec
1
Frequency : f = (# cycles / sec), [ f ] = Hz
Τ
2π
Angular Frequency : ω = 2π f =
, [ω ] = rad / s
Τ
Displacement
of Mass
Simple Harmonic Motion
x ( t ) = A cos θ ( t )
2π
θ (t ) = ω t =
t
Τ
Displacement
of Mass
Simple Harmonic Motion
x ( t ) = A cos ω t
v(t ) = − Aω sin ω t
2
a (t ) = − Aω cos ω t
Displacement
of Mass
Notice:
a = −ω x
2
2π
ω=
Τ
Newton’s 2nd Law?
x ( t ) = A cos ω t
F
=
ma
v(t ) = − Aω sin ω t
2
= m(−ω x)
a (t ) = − Aω cos ω t
2
a = −ω 2 x
= −kx
k = mω
k
ω=
m
2
angular
frequency
2π
ω=
Τ
Simple Harmonic Motion
k
ω=
m
Does the period
depend on the
displacement, x?
T=
2π
ω
m
T = 2π
k
The period
depends only
on how stiff the
spring is and
how much
inertia there is.
Finding The Spring Constant
A 2.00 kg block is at rest at the end of a horizontal spring on a
frictionless surface. The block is pulled out 20.0 cm and released.
The block is measured to have a frequency of 4.00 hertz.
a) What is the spring constant?
1 ω
1
f = =
=
Τ 2π 2π
k
m
k = 4π 2 f 2 m
2
m
k = 4π 2 (4 / s) 2 (2kg ) 2 = 1263 N / m
m
Harmonic Motion Described
Frestoring = − kx
m
k
T = 2π
, ω=
k
m
x(t ) = A cos ω t
v(t ) = − Aω sin ω t
2
a (t ) = − Aω cos ω t
Maximum Velocity and Acceleration
v = − Aω sin ω t
vmax = Aω
@ω t = nodd
π
2
a = − Aω cos ω t
2
amax = Aω
@ ω t = nπ
2
Finding Maximum Velocity
A 2.00 kg block is at rest at the end of a horizontal spring on a
frictionless surface. The block is pulled out 20.0 cm and released.
The block is measured to have a frequency of 4.00 hertz.
a) What is the spring constant? b) What is the maximum velocity?
k = 1263 N / m
v(t ) = − Aω sin ω t
vmax
vmax = Aω
1263N / m
= (.2m)
2kg
k
=A
m
= 5.0m / s
Simple Harmonic Motion
Consider a m = 2.00 kg object on a spring in a horizontal
frictionless plane. The sinusoidal graph represents displacement
from equilibrium position as a function of time.
A) What is the amplitude of motion?
B) What is the period, T? frequency, f ?
C) What is the equation representing the displacement?
A) A = . 08 m
B) T = 4s
f = 1/T = .25Hz
2π
2π
π
t = .08m cos
t = .08m cos t
C) x(t ) = A cos
2
4s
Τ
Simple Harmonic Motion
Consider a m = 2.00 kg object on a spring in a horizontal plane.
The sinusoidal graph represents displacement from equilibrium
position as a function of time. x(t ) = .08m cos π t
2
D) What is the speed of the object at t = 1s?
v = − Aω sin ω t
vmax
2π
2π
vmax = − Aω = − A
= −.08m
Τ
4s
= −.13m / s
E) What is the acceleration of the object at t = 1s?
zero!
Simple Harmonic Motion
Consider a m = 2.00 kg object on a spring in a horizontal plane.
The sinusoidal graph represents displacement from equilibrium
position as a function of time. x(t ) = .08m cos π t
2
F) What is the spring constant?
m
T = 2π
k
4π 2 m
k=
Τ2
k = 4.94 N / m
4π 2 (2kg )
=
(4 s ) 2
SHM Question
The position of a simple harmonic oscillator is given by
⎛π ⎞
x (t ) = (0.5 m) cos ⎜ t ⎟
⎝3 ⎠
where t is in seconds.
a) What is the period of this oscillator ?
b) What is the maximum velocity of this oscillator?
a)
b)
x(t ) = A cos ω t
π
2π
=ω =
3
Τ
v(t ) = − Aω sin ω t
vmax
2π
= 0.5m
6s
vmax = Aω
= 0.52m / s
Τ = 6s
2π
=A
Τ
Energy in a Spring
Mechanical Energy is Conserved!
PEspring
1 2
= kA
2
KEspring
1 2
= mv
2
Energy in a Spring: Quiz Question
What speed will a 25g ball be shot out of a toy gun if the spring
(spring constant = 50.0N/m) compressed 0.15m? Ignore friction.
Use Energy!
PEspring = KEball
1 2 1 2
kx = mv
2
2
k
v=
x
m
Note: this is also
=ωA
= vmax
50.0 N / m
v=
(.15m) = 6.7 m / s
.025kg
Spring Question
A 1.0-kg object is suspended
from a spring with k = 16 N/m.
The mass is pulled 0.25 m
downward from its equilibrium
position and allowed to oscillate.
What is the maximum kinetic
energy of the object?
PEg = 0
Maximum kinetic energy happens at the equilibrium position.
KEmax
v(t ) = − Aω sin ω t
1 2
= mvmax
2
vmax = Aω
2
KEmax
k
=A
m
1 ⎛
k ⎞
1 2
1 2
= mvmax = m ⎜⎜ A
⎟⎟ = A k = 0.5J = PEm ax
m⎠
2 ⎝
2
2
Simple Pendulum
For small angles, simple pendulums exhibit Simple
Harmonic Motion:
Simple Pendulum
For small angles, simple pendulums exhibit Simple
Harmonic Motion:
Restoring Force:
F = −mg sin θ ≈ −mgθ
θ = s/L
F ≈ −mgs / L = −ks
k = mg / L
Amplitude : x = s
For angles less than 15 degrees:
ω = k m = mg / L m = g L
ω=
g
L
L
T = 2π
g
Moon Clock
At what rate will a pendulum clock run on
the moon where gM = 1.6m/s2, in hours?
Τ Earth
L
= 2π
gE
Τ Moon
L
= 2π
gM
Divide:
ΤMoon
=
Τ Earth
L
2π
gM
L
2π
gE
Τ Moon = Τ Earth
gE
= 2.45h
gM
•
•
•
•
result from periodic disturbance
same period (frequency) as source
Longitudinal or Transverse Waves
Characterized by
1
f =
Τ
– amplitude (how far do the “bits” move from their
equilibrium positions? Amplitude of MEDIUM)
– period or frequency (how long does it take for each “bit” to
go through one cycle?)
– wavelength (over what distance does the cycle repeat in a
freeze frame?)
f
– wave speed (how fast is the energy transferred?) v =
λ
Types of Waves
Sound
String
Sound is a Longitudinal Wave
Pulse
Tuning Fork
Guitar String
Spherical Waves
Wavelength and Frequency are Inversely related:
f =
The shorter the wavelength, the higher the frequency.
The longer the wavelength, the lower the frequency.
3Hz
5Hz
v
λ
Wave speed: Depends on Properties of the Medium:
Temperature, Density, Elasticity, Tension, Relative Motion
v=λf
Speed of wave depends on
properties of the MEDIUM
v=λf
Speed of particle in the
Medium depends on
SOURCE: SHM
v(t ) = − Aω sin ω t
Waves on Strings
v=λf
v=
F
μ
(1D string)
μ = m / L (linear mass density)
Problem:
v=λf
The displacement of a vibrating string vs position along the
string is shown. The wave speed is 10cm/s.
A) What is the amplitude of the wave?
B) What is the wavelength of the wave?
C) What is the frequency of the wave?
4cm
6cm
1.67 Hz
10cm / s
f = v/λ =
= 1.67 Hz
6cm
Problem:
v=λf
The displacement of a vibrating string vs position along the
string is shown. The wave speed is 10cm/s.
D) If the linear density of the string is .01kg/m, what is the
tension of the string?
Problem:
F
v=
m/ L
The displacement of a vibrating string vs position along the
string is shown. The wave speed is 10cm/s.
D) If the linear density of the string is .01kg/m, what is the
tension of the string?
F = v (m / L)
2
−5
F = (.1m) (.01kg / m) = 10 N
2
Problem:
F
v=
m/ L
The displacement of a vibrating string vs position along the
string is shown. The wave speed is 10cm/s.
e) If the the tension doubles, how does the wave speed change?
Frequency? Wavelength?
v2 =
F2
2F
=
= 2v
m/ L
m/ L
Wave speed increases by a factor of
2
Wave PULSE:
•
•
•
•
traveling disturbance
transfers energy and momentum
no bulk motion of the medium
comes in two flavors
• LONGitudinal
• TRANSverse
Reflected PULSE:
If the end is bound, the pulse
undergoes an inversion upon
reflection: “a 180 degree
phase shift”
If it is unbound, it is not
shifted upon reflection.
Free End
Bound End
Reflection of a Wave Pulse
Reflection of a Traveling Wave
Superposition
Waves interfere temporarily.
Pulsed Interference
Standing Waves: Boundary
Conditions
Transverse Standing Wave
Produced by the superposition of two identical
waves moving in opposite directions.
Standing Waves on a String
Harmonics
Standing Waves on a String
λ1 = 2L
λ2 = L
2L
λ3 =
3
Standing Waves on a String
2L
λn =
n
f n = v / λn
v
fn = n
2L
Standing Waves on a String
Harmonics
The possible frequency and energy states of a wave on a string
are quantized.
v
v
f1 = =
λ 2l
v
fn = n
2l
f n = nf1
Strings & Atoms are Quantized
The possible frequency and energy states of an electron in an
atomic orbit or of a wave on a string are quantized.
v
f =n
2l
En = nhf , n= 0,1,2,3,...
h = 6.626 x10
−34
Js
Multiple Harmonics can be present at the same time.
Which harmonics (modes) are present on the string?
The Fundamental and third harmonic.
Standing Waves
Standing waves
form in certain
MODES based on
the length of the
string or tube or
the shape of drum
or wire. Not all
frequencies are
permitted!