Transcript these slides - Department of Physics | Oregon State

Displacement of an Object in Simple Harmonic Motion
x(t) = A·cos(ωt + f0)
x is the displacement (in ±meters) at a time t. x is a vector.
A is the maximum displacement (amplitude) magnitude of the cosine
function, expressed in meters. Since cosine varies between –1 and
+1, x will oscillate between – A and +A. Those extrema occur
whenever (ωt + f0) = nπ, where n = 0, 1, 2, 3, …
ω is the angular frequency, in radians/second. The angular
frequency is related to frequency and period: ω = 2πf = 2π/T
Note that ω does not have units of hertz (cycles per second).
t is the time (in seconds).
f0 is the “angular position” of the oscillator in its cycle at t = 0.
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OSU PH 212, Before Class #21
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Velocity of an Object in Simple Harmonic Motion
Since vx(t) = dx/dt:
vx(t) = – ωA·sin(ωt + f0)
vx is the x-velocity at a time t
ωA is the maximum speed, vmax, which is the amplitude of this
velocity sine function; since sine varies between –1 and +1, vx will
vary between – ωA and + ωA.
Those maxima occur at (ωt) = nπ/2, where n = 1, 3, 5,….
Note that maximum speed occurs at the minimum displacement; and
minimum speed occurs at the maximum displacement.
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Acceleration of an Object in Simple Harmonic Motion
Since ax(t) = dv/dt:
ax(t) = –ω2A·cos(ωt + f0)
ax is the x-acceleration at a time t
ω2A is the maximum acceleration; cosine varies between –1 and +1,
so ax varies between –ω2A and + ω2A.
Those maxima occur at (ωt) = nπ, where n = 0, 1, 2, 3, ….
Note: Maximum acceleration occurs at maximum displacement (and
minimum velocity). And minimum acceleration (zero!) occurs at
minimum displacement (zero!) where velocity is maximum.
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Review: A mass is moving at constant speed in a circle of radius A
meters. It completes each revolution in a time of T seconds. Its
speed and acceleration magnitudes are these:
v = A
a = A2
Q: How does this relate to amax and vmax of a mass oscillating on a
spring?
A: Either axis of the 2D circular motion exhibits the above
behavior—which matches that of a spring/mass system,
where, by Newton’s 2nd Law, amax = Fmax/m = kA/m.
And, by energy considerations: (1/2)mvmax2 = (1/2)kA2
Each of these brings us the same conclusion:  = √(k/m)
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Frequency of an Object in Simple Harmonic Motion
If no external work is done on the spring-mass system, its total
mechanical energy remains constant throughout its oscillation:
Emech.total = KT + Uspr = (1/2)mv2 + (1/2)kx2
And we know something in particular about Emech.total at certain
points in the oscillation cycle:
Where all the energy is in the form of KT (at x = 0),
Emech.total = (1/2)mvmax2 = (1/2)m[ωA]2
Where all the energy is in the form of Uspr (at x = ±A),
Emech.total = (1/2)kA2
Thus: (1/2)kA2 = (1/2)m[ωA]2
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Or:
OSU PH 212, Before Class #21
ω = √(k/m)
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Example: Four springs are compressed from their equilibrium
position at x = 0. When released, they will oscillate. Rank in order,
highest to lowest, their maximum speeds.
A. c > b > a > d
B. c > b > a = d
C. a = d > b > c
D. d > a > b > c
E. b > c > a = d
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Example: Four springs are compressed from their equilibrium
position at x = 0. When released, they will oscillate. Rank in order,
highest to lowest, their maximum speeds.
A. c > b > a > d
B. c > b > a = d
C. a = d > b > c
D. d > a > b > c
E. b > c > a = d
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More good examples: Try exercises 17 and 19 on page 415.
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The sinusoidal nature of simple harmonic motion (SHM) was first
deduced by observation—comparing it to uniform circular motion.
But the equation of motion, x(t) = Acos(ωt + f0), also emerges
directly from Newton’s 2nd Law of motion for a mass on a spring:
Fx.net(t) = max(t)
–kx(t) = max(t)
–kx(t) = m(d2x/dt2)
That is: d2x/dt2 = –(k/m)x(t)
The solution to any differential equation of this form (where the
function’s second derivative is proportional to the function) is a
sinusoidal (sine or a cosine) function. And the constant of
proportionality is –ω2, where ω is the angular frequency of the
resulting motion. And we see oscillations described by this all over
nature…
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The motion of a mass hanging vertically from an ideal spring
• Where is the equilibrium position?
• Is the velocity constant?
• Is the speed constant?
• Where is the mass moving the fastest? The slowest?
• Where is the force the largest? The smallest?
• Where is the acceleration the largest? The smallest?
• What is the net y-force on the mass?
• What is the equation of motion, y(t)?
(See pp. 402-403 for the full derivation.)
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The motion of a simple pendulum
• Where is the equilibrium position?
• Is the velocity constant?
• Is the speed constant?
• Where is the mass moving the fastest? The slowest?
• Where is the force the largest? The smallest?
• Where is the acceleration the largest? The smallest?
The key to answering all these questions is to recognize the familiar
form of the equation of motion (and see also pp 404-405).…
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For a simple pendulum:
Fnet = –mg·sin θ = mat
at = –g·sin θ
But at(t) = d2s/dt2, so:
d2s/dt2 = –g·sin θ
Small angle approximation: For angles less than about 10 degrees
(about 0.17 rad), sin θ ≈ θ ≈ s/L.
d2s/dt2 = –(g/L)s
So:
with a solution of s(t) = Acos(ωt + f0), where
So the frequency, f, for a small-angle, simple pendulum is given by
f = [√(g/L)/(2p)]
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Likewise, consider now the physical pendulum. This is a pendulum
whose mass, M, is not all located only at the end of the arm. Its
resistance to angular acceleration about the pivot is described by its
moment of inertia, I. And its center of mass is located a distance l
from the pivot. Therefore:
tnet = trestoring = –Mgl·sin θ = Ia
a = –g·sinθ
But a(t) = d2θ/dt2, so:
d2θ/dt2 = –(Mgl/I)·sinθ
Again, for small angles, sinθ ≈ θ.
d2θ/dt2 = –(Mgl/I)θ
So:
with a solution of θ(t) = Acos(ωt + f0), where ω = √[Mgl/I]
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