springs & simple harmonic motion
TRANSCRIPT
Springs & Simple Harmonic Motion
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• A bird on a perch can put itself into a
periodic oscillation.
• An oscillation is a back and forth
motion over the same path.
• It is said to be periodic if each cycle of the motion takes place in equal periods of time.
• Simple Harmonic Motion (SHM) can be described as a projection of circular motion on one axis.
SHM and Circular Motion• An object moving with constant speed in a
circular path observed from a distant point will appear to be oscillating with simple harmonic motion.
• The shadow of a pendulum bob moves with s.h.m. when the pendulum itself is either oscillating or moving in a circle with constant speed.
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SHM and Circular MotionFor any SHM there is a corresponding circular
motion.
• the radius of the circle is equal to the amplitude of the SHM
• the time period of the circular motion is equal to the time period of the SHM
• The relationship of circular
motion and SHM is
a
rT
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Cutnell & Johnson, Wiley Publishing,
Physics 5th Ed.
SHM and Circular Motion• Rearranging,
• Recall, therefore• The constant of proportionality between acceleration
and displacement for an object moving with s.h.m. is equal to the square of the angular velocity of the corresponding circular motion.
• So, to find the value of the constant for a given oscillation, we simply measure the time period and then use the relation
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Spring ReviewHooke’s Law• Fs=restoring force of a spring• k = spring constant• x = displacement of the spring
• The friction free motion shown above is known as Simple Harmonic Motion (SHM)
Fs=-kx
Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.
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Spring ReviewSHM graph
• Sinusoidal motion
• max. stretching distance (x) from the equilibrium position is equal to the amplitude of the graph.
Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.
Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.
SHM has displacement, velocity and acceleration.
Displacement:
Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.
SHM and Circular Motion
sinrx t trx sin
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Note: r = amplitude (A)
SHM and Circular MotionPeriod (T) – time required to complete one cycle.
• Recall, for 1 cycle &
• Therefore,
Frequency (f) – number of cycles
per second
• Units – Hertz (Hz)
t
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T
2
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Tf
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SHM Terms
Copywrited by Holt, Rinehart, & Winston
SHM and Circular MotionVelocity
• Velocity of the shadow is the vx of vT.
Recall
At x =0 m v vmax
Therefore, θ = 0°
or
t rvT
cosTvv
trv cos
rv max
rv max
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SHM and Circular MotionAcceleration
• Acceleration of the shadow is the ax of ac.
Recall
At x = r, a amax
Therefore, θ = 90°
or
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Note: r = amplitude (A)
Graphs Describing S.H.M.• Displacement against time x = rsin(ωt)
• Velocity against time v = rωcos(ωt)
• Acceleration against time a = -rω²sin(ωt)
Note: All these graphs assume that, at t = 0, the body is at the equilibrium position.
SHM EnergyIn SHM the total energy possessed by the
oscillating body does not change with time.
Recall, Total Mechanical Energy = Kinetic Energy + Potential Energy
• An oscillation in which the total energy decreases with time is described as a damped oscillation.– Due to air resistance or other similar causes
SHM K.E. & P.E.• Kinetic energy against time ½m[rωcos(ωt)]²
• Since, P.E. = T.M.E. - K.E.
• P.E. graph has the same form as the K.E. graph but is inverted.
SHM K.E. & P.E.
When,
• P.E. is a maximum, K.E. is a minimum (K.E.=0)
• K.E. is a maximum, P.E. is a minimum (P.E.=0)
T.M.E = P.E. + K.E.
http://ecommons.uwinnipeg.ca/archive/00000030/02/shmani2.gif
http://www.maths-physics.nuigalway.ie/Maple_animations/images%5CSHM13.gif
Copywrited by Holt, Rinehart, & Winston
SHM of SpringsDerivation of Frequency of SHM
• Hooke’s Law: F=-kx
• F = ma = -kx
• Frequency,
• Period,
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Elastic Potential Energy (PEelastic)
• Energy that a spring contains by being stretched or compressed.
221 kxPEelastic
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
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Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.
Total Mechanical EnergyWhat forms of Mechanical Energy have we
discussed?
• Combining of all these
elasticalgravitaionrotationalnaltranslatio PEPEKEKETME
2212
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21 kxmghImvE
SHM of Springs
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Energy ProblemAn object of mass m = 0.200 kg is vibrating on a
horizontal frictionless table as shown. The spring has a spring constant k = 545 N/m. It is stretched initially to xo =4.50 cm and released from rest. Determine the final translational speed vf of the object when the final displacement of the spring is (a) xf = 2.25 cm and (b) xf=0 cm.
Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.
The Simple Pendulum • When displaced from its
equilibrium position by an angle θ and released it swings back and forth.
• Plotting the motion reveals a pattern similar to the sinusoidal motion of SHM
Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.
Copywrited by Holt, Rinehart, & Winston
http://www.enm.bris.ac.uk/teaching/pendulum/animations/timeseries_simple.gif
The Simple Pendulum • The gravitational force (Fgx)
provides the torque.
• This restoring force (Fgx):
• Since the displacement and restoring force act in opposite directions.
• The torque of the pendulum is
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The Simple Pendulum For small angles (10°or smaller)
• θ = sinθ
• Where k‘ is a constant independent of θ
• This form is similar to F=-kx (Hooke’s Law)
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The Simple Pendulum For small angles ONLY
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Energy of a PendulumTME = Constant
TME = PE + KE
PEmax KE = 0
KEmax PE = 0
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Swinging ProblemDetermine the length of a simple pendulum that
will swing back and forth in SHM with a period of 1.00 s.
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Simple Harmonic Motion
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Elastic Deformation• When a spring is stretched and released it
returns to its original shape.
• Likewise, some materials when stretched or compressed and released return to their original shape
Elastic Materials
• Explained, by modeling the chemical bonds between as atoms as springs. When force causing deformation is removed, material returns to original shape
Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.
Elasticity• A property of a body that causes it to deform
when a force is exerted and return to its original shape when the deforming force is removed, within certain limits
Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.
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Stress ( )• The force exerted on an area divided by the
area
• Units Newton per square meter (N/m2 )
or pascal (Pa)
A
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Strain ( )• The resulting fractional change in length (ΔL/Lo)
due to a stretch/compression deformation.
• unitless
lengthoriginal
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oL
L
Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.
Young’s Modulus (Y)• Measure of the elasticity of a material.
• Ratio of stress to strain of a material.
• A material property
• SI units Newton per square meter (N/m2 )
Ystrain
stressModulussYoung '
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Shear Deformation• Change in shape of an object due to the
application of equal parallel forces in opposite directions.
Shear Stress ( )
Shear Strain ( )
oL
x
http://scign.jpl.nasa.gov/learn/plate5.htm
Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.
Shear Modulus (G)• Ratio of shear stress to shear strain of a
material.
• A material property
• SI units Newton per square meter (N/m2 )
strainshear
stressshearModulusShear
G
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Hooke’s Law• Stress is directly proportional to strain
– Slope equal to Young’s modulus or Shear Modulus depending on measurement
• Elastic region
Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.
Stress Strain CurveProportionality limit – point on a stress strain
curve where stress and strain are no longer directly proportional.
Elastic limit – point above which the material will not return to its original shape– Above inelastic region
Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.
Elastic Properties of Selected Engineering Materials
Material Density(kg/m3)
Young's Modulus109 N/m2
Ultimate Strength Su
106 N/m2
Yield Strength Sy
106 N/m2
Steela 7860 200 400 250
Aluminum 2710 70 110 95
Glass 2190 65 50b ...
Concretec 2320 30 40b ...
Woodd 525 13 50b ...
Bone 1900 9b 170b ...
Polystyrene 1050 3 48 ...
a Structural steel (ASTM-A36), b In compression, c High strength, d Douglas firData from Table 13-1, Halliday, Resnick, Walker, 5th Ed. Extended.
Applications:NONLINEAR CONSTITUTIVE BEHAVIOR OF PZT
-160
-140
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-100
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-60
-40
-20
0
-0.8 -0.6 -0.4 -0.2 0S
tre
ss (
MP
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Strain (%)
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-20
0
-0.7 -0.6 -0.5 -0.4 -0.3 -0.2 -0.1 0
Str
ess
(M
Pa
)
Strain (%)
Setup
Loading Fixture
Alumina Block
Electrode
Specimen
Oil Bath Fixture
Tilt Table
Adjustable Screws
Alumina Spacer
Brass Electrode
Grafoil
PZT Sample (electroded surface)
Oil Bath
Strain GagesGrafoil
Brass Electrode
Alumina Spacer
Force
Force
Setup
+- 5 V
C=10 F-
+
+ 2000
20 KV Power Amplifier
Function Generator
Electrometer
Vout
ground
D = Q/A = C(Vout)/A
+
-
Strain Gage Amplifier
Force
Load Cell
ADC
Electrodes
Macroscopic Depth-Sensing Indentation
TestsWill Stoll, Physics Teacher, Norcross High School
International Site Mentor: Dr. T. H. Zhang and Professor F. J. Ke, Institute of Mechanics Beijing, China
Georgia Tech Mentor: Professor Min Zhou
Macroscopic Depth Sensing Indentation (DSI) Tests
• A Vicker indenter is inserted into the surface of the specimen.
• The load (P) and the displacement (h) of the indenter into the specimen are measured and plotted in a P vs h curve.
• Material properties of hardness (H) and modulus of elasticity (E) can be calculated from the P vs h curve.
Research Focus• Well established test for the nano-scale and
micro-scale regions.
• Investigate the applicability of DSI to the macro-scale region.
• Determine whether DSI testing is independent of scale.
•Hardness (H) of a specimen is calculated from:
where Ac is the contact area which is a function of the
displacement (h) of the indenter into the specimen .
•The reduced modulus Er is derived from the relationship
where S is the stiffness which is dependent on the unloading curve.
Background
2r
c
SE
A
max
c
PH
A
(The reduced modulus is used to account for the elastic deformation in both the indenter and specimen)
Test Setup• Load applied with a 250 N
load cell attached to an Instron Microtester
• Vicker’s Indenter attached to piston connected to the load cell.
• Specimen glued to attached base plate of test frame.
Instron 5848 Mirotester
frame
Load cell
DWS
Base jig
Upper jig
Indenter
Specimen
Wing
Piston
• Used a displacement capacitance sensor
• Decouples displacement from the load cell eliminating frame compliance.
Sensor
Capacitance gauge
Signal process and show
Target
FIG. 4. Measurement schematic of the capacitance displacement sensor
Independent Displacement Measurement
Loading Profile• Load tests of 2.5 N, 5 N, 10 N, 25 N, 50 N, 100
N, 150 N and 200 N performed.• Slow ramp to maximum load by controlling
displacement rate– For 2.5 N, 5 N and 10 N tests - displacement rate of 0.1 m/s. – For 25 N, 50 N tests - displacement rate of 0.5 m/s.– For 100 N, 150 N and 200 N tests - displacement rate of 1.0
m/s.
• Maximum load held for 30 seconds• Unloaded to 90% of the maximum load at a
constant rate in 80 seconds
Analysis• Hardness calculated by assuming ideal
indenter tip area. – ISO 14577 standard recommends this for
depths greater than 6 m
• Modulus calculated by finding stiffness which is the slope of the unloading curve.– Power law fit used for top 50% of unloading
curve.
Test Results
0 20 40 60 80 100
0
50
100
150
200
L
oa
d ()
Depth (m)
200N 150N 100N 50N 25N 10N 5N 2.5N
Hardness and Modulus Results
Peak Load (N)
Hardness (GPa)
Modulus (GPa)200 1.09 68.4150 1.06 67.8100 1.18 73.050 1.14 64.925 1.12 65.210 1.16 69.45 1.19 70.22.5 1.18 81.7
Aluminum accepted values: H = 1.1 GPa E = 71 GPa
Testing Challenges• Non-uniform adhesive layer
– Change mounting of specimen
• Load cell drift– Complicates calculation of the zero point– After unloading include a holding period to establish
the drift rate and then correct the displacement data for this
Conclusion• Results reinforce the validity of macroscopic DSI tests• DSI tests allow the hardness and modulus to be
determined without direct measurements of the contact area at different scale lengths.
• DSI testing holds great promise yet some refinement is still needed before widespread use.