Phys 2101
Gabriela González
Oscillations = anything that “swings” back and forth.
Harmonic oscillations: described with trigonometric
functions.
Three kinds of oscillations:
•  free,
•  damped and
•  forced.
We will focus on free oscillations.
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Displacement:
angular frequency
phase angle
x(t) = xm cos(ωm t + φ)
maximum (peak) amplitude
frequency (cps, Hz): f=ω/2π
period (sec) : T=1/f=2π/ω
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Displacement:
x(t) = xm cos(ω t + φ)
Velocity:
v(t) = dx(t)/dt = -ω xm sin(ω t + φ)
Acceleration:
a(t) = dv(t)/dt = -ω2 xm cos(ω t + φ) = - ω2 x(t)
Force:
F = m a = -m ω2 x = -k x
A spring! ω2=k/m
A spring force produces oscillations of the mass attached to it.
Any oscillation has a “spring-like” (restoring) force.
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Displacement:
x(t) = xm cos(ω t + φ)
Velocity:
v(t) = dx(t)/dt = -ω xm sin(ω t + φ)
Acceleration:
a(t) = dv(t)/dt = -ω2 xm cos(ω t + φ)
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A 5.00 kg block is attached to a spring with spring constant k=2.50 N/cm.
If the spring is then stretched 12.0 cm and released from rest,
a)  what is its period of oscillation?
b)  what is its maximum amplitude of oscillation?
c)  what is its maximum speed?
d)  what is the maximum force the spring will exert on the body?
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A 5.00 kg block is attached to a spring with spring constant k=2.50 N/cm.
If the spring is then stretched 12.0 cm and released from rest,
a)  what is its period of oscillation?
b)  what is its maximum amplitude of oscillation?
c)  what is its maximum speed?
d)  what is the maximum force the spring will exert on the body?
a)  Angular frequency is ω = (k/m)½ = (250 N/m /5 kg) ½ = 7.1 rad/sec.
Period is T = 2π/ω = 0.89 sec
b)  The maximum stretching of the spring is 12cm, so the amplitude is xm= 0.12m.
c)  The maximum speed is vm = ωxm = 0.85 m/s.
d)  The maximum spring force is Fm= kxm = 30 N.
Or, maximum acceleration is am = ω2xm = 60.5 m/s2, and Fm= mam = 30N.
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A 5.00 kg block is attached to a spring with spring constant k=2.50 N/cm.
If the spring is then stretched 12.0 cm and released from rest,
How do frequency, period, amplitude, maximum speed and force change if …
1.  … the mass is heavier?
2.  … the spring is stiffer?
3.  … the initial displacement is larger?
4.  … the mass is pushed when releasing it?
What answers (for period, amplitude, maximum speed and force) would change if …
1.  … the mass was heavier?
Frequency is lower, period is longer, amplitude is the same, max speed is smaller, max force is smaller.
2.  … the spring was stiffer?
Frequency is higher, period is shorter, amplitude is the same, max speed is larger, max force is larger.
3.  … the initial displacement was larger?
Frequency and period are the same, amplitude is larger, max speed is smaller, max force is smaller.
4.  … the mass is pushed when releasing it?
The frequency and period are the same, amplitude is larger, max speed and force are larger.
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A block that weights 50N hangs from a spring,
extending it 20cm from its unstretched position.
•  What is the spring constant?
The block is removed and a 5 kg body is hung from the same
spring, letting it reach equilibrium. If the spring is then
stretched 12.0 cm further and released from rest,
•  what is its period of oscillation?
•  what is its maximum amplitude of oscillation?
•  what is its maximum speed?
•  what is the maximum force the spring will exert on the body?
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If F=-kx, potential energy = -work done by the force is U=½ k x2.
Total mechanical energy is potential plus kinetic:
E= U + K
= ½ k x2 + ½ m v2
= ½ k (xm cos(ωt +φ))2 + ½ m (-ωxm sin(ωt +φ))2
= ½ k xm2 cos2(ωt +φ) + ½ m ω2 xm2 sin2(ωt +φ)
= ½ k xm2 cos2(ωt +φ) + ½ m (k/m) xm2 sin2(ωt +φ)
=½ k xm2 cos2(ωt +φ) + ½ k xm2 sin2(ωt +φ)
= ½ k xm2 (cos2(ωt +φ) + sin2(ωt +φ))
= ½ k xm2
Total mechanical energy is constant, because only conservative forces are acting.
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The block has a kinetic energy of 3J and the spring an elastic
potential energy of 2 J when the block is at x=2.0cm.
(a)  What is the kinetic energy at x=0?
(b)  What is the potential energy at x=-2.0cm?
(c)  What is the potential energy at x=-xm?
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Restoring torque (“spring”): τ=-κθ
The angular displacement follows a simple harmonic motion:
θ(t)=θm cos(ω t + φ)
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with ω = κ/I (I= rotational inertia).
Period: T= 2π (I/κ) ½ An engineer has an odd shaped object and
needs to find its rotational inertia about an
axis through the center of mass. She has a
wire of unknown properties, a rod, a
measuring tape, a stopwatch, and a scale.
How can torsion pendulums help her?
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The horizontal force on a simple pendulum is F = -mg sin θ
For small oscillations, sin θ~ θ=s/L, so
F = -(mg/L) s : another spring!
SHM:
s = sm cos(ω t + φ)
with ω2=(mg/L/m)=g/L
and period T= 2π (L/g) ½
Period is independent of mass!!
A geophysicist is asked to measure how
much the acceleration of gravity changes at
different points in a mountain. He takes with
him just a mass, a string, a measuring tape,
and a stopwatch. How does he manage?
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Restoring torque is τ = -mgh θ
Angular frequency is ω2=mgh/I
and period is T=2π (I/mgh) ½
Is a simple pendulum also a
physical pendulum?
Which pendulum has a longer period?
  a stick of length L and mass M
  a mass M at the end of a string of length L
  a disk with diameter L and mass M, hanging from its edge.
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