Physics 101: Lecture 20 Elasticity and Oscillations

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Physics 101 Lecture 20 Elasticity and
Oscillations

• Todays lecture will cover Textbook Chapter
  10.5-10.10

• Exam II tonight
• No discussion quiz this week
• Labs start again on Wed

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Review Energy in SHM

• A mass is attached to a spring is set in motion
  by stretching to a maximum displacement xA and
  releasing
• Energy U K constant!
• ½ k x2 ½ m v2
• At maximum displacement xA, v 0
• Energy ½ k A2 0
• At zero displacement x 0
• Energy 0 ½ mvm2
• ½ k A2 ½ m vm2
• vm ?(k/m) A
• Analogy with gravity/ball

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Kinetic Energy ACT

• In Case 1 a mass on a spring oscillates back and
  forth. In Case 2, the mass is doubled but the
  spring and the amplitude of the oscillation are
  the same as in Case 1. In which case is the
  maximum kinetic energy of the mass bigger?
• A. Case 1
• B. Case 2
• C. Same

½kA2 ½mvm2
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Potential Energy ACT

• In Case 1 a mass on a spring oscillates back and
  forth. In Case 2, the mass is doubled but the
  spring and the amplitude of the oscillation are
  the same as in Case 1. In which case is the
  maximum potential energy of the mass and spring
  bigger?
• A. Case 1
• B. Case 2
• C. Same

Maximum displacement x A Energy ½ k A2 0
Same for both!
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Velocity ACT

• In Case 1 a mass on a spring oscillates back and
  forth. In Case 2, the mass is doubled but the
  spring and the amplitude of the oscillation are
  the same as in Case 1. Which case has the larger
  maximum velocity?
• 1. Case 12. Case 23. Same

Same maximum Kinetic Energy K ½ m v2
smaller mass requires larger v
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Simple Harmonic MotionQuick Review
x(t) Acos(?t) v(t) -A?sin(?t) a(t)
-A?2cos(?t)
x(t) Asin(?t) v(t) A?cos(?t) a(t)
-A?2sin(?t)
OR
Period T (seconds per cycle) Frequency f
1/T (cycles per second) Angular frequency ?
2?f 2?/T
xmax A vmax A? amax A?2
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Natural Period T of a Spring

• Simple Harmonic Oscillator
• w 2 p f 2 p / T
• x(t) A cos(wt)
• v(t) -Aw sin(wt)
• a(t) -Aw2 cos(wt)
• Draw FBD write Fma
• -k x m a
• -k A m amax
• -k A m (-A w2)
• Aw2 (k/m) A
• w ?(k/m)

A,m,k dependence demo
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Period ACT

• If the amplitude of the oscillation (same block
  and same spring) is doubled, how would the period
  of the oscillation change? (The period is the
  time it takes to make one complete oscillation)
• A. The period of the oscillation would double.B.
  The period of the oscillation would be halvedC.
  The period of the oscillation would stay the same

x
2A
t
-2A
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Equilibrium position and gravity

• If we include gravity, there are two forces
  acting on mass. With mass, new equilibrium
  position has spring stretched d
• SFy 0
• kd mg 0
• d mg/k Let this point be y0
• SF ma
• k(d-y) mg ma
• -k y ma
• Same as horizontal! SHO
• New equilibrium position y0
• corresponds to height -d

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Vertical Spring ACT

• Two springs with the same k but different
  equilibrium positions are stretched the same
  distance A and then released. Which would have
  the larger maximum kinetic energy?
• 1) M 2) 2M 3) Same

k
k
Y0
Just before being released, v0 yA Etot 0 ½
k A2 Same total energy for both When pass
through equilibrium all of this energy will be
kinetic energy again - same for both!
Y0
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Pendulum Motion
Bowling ball pendulum

• For small angles
• T mg cos(q) mg
• Tx -mg sin(q) -mg x/L Note F
  proportional to x!
• S Fx m ax
• -mg (x/L) m ax
• ax -(g/L) x
• Recall for SHO a - w2 x
• w ?(g/L)
• T 2 p ?(L/g)
• Period does not depend on A, or m!

q
L
T
m
x
mg
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Preflight 1

• Suppose a grandfather clock (a simple pendulum)
  runs slow. In order to make it run on time you
  should
• 1. Make the pendulum shorter
• 2. Make the pendulum longer

82 18
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Elevator ACT

• A pendulum is hanging vertically from the ceiling
  of an elevator. Initially the elevator is at
  rest and the period of the pendulum is T. Now
  the pendulum accelerates upward. The period of
  the pendulum will now be
• A. greater than T
• B. equal to T
• C. less than T

g is effectively bigger, T is lower.
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Preflight

• Imagine you have been kidnapped by space invaders
  and are being held prisoner in a room with no
  windows. All you have is a cheap digital
  wristwatch and a pair of shoes (including
  shoelaces of known length). Explain how you might
  figure out whether this room is on the earth or
  on the moon

Attach the shoe to the lace and swing the shoe
like a pendulum. Time the period. Multiply 2pi by
the sq. root of (Length of lace/9.8). If this is
equal to the measured period then you are on
Earth. If not then the gravitational force is
different and you are on the moon.
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Alien Preflight
Jump and see the effect of gravity.
I would drop my shoes from a height I measured
with my shoelaces and time how long it takes them
to hit the floor. I would then calculate the
acceleration from our kinematics equations, and
if a9.8 I was on the Earth, but if a was much
less than 9.8 I was on the moon.
I would jump into the air and see if I landed in
a normal time or if I felt like I weighted much
less then normal. But if the aliens brainwashed
me and I could not remember normal common sense,
only physics, then i would make a pendulum with
my shoes and shoelaces, and time the oscillation
period of the shoes for two different lengths of
string. The period of the pendulum swing is equal
to (2pi)(L/g)0.5, so on the earth, where g is
larger, a change in L will have less of an effect
than on the moon, where g is much smaller.
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Summary

• Simple Harmonic Motion
• Occurs when have linear restoring force F -kx
• x(t) A cos(wt)
• v(t) -Aw sin(wt)
• a(t) -Aw2 cos(wt)
• Springs
• F -kx
• U ½ k x2
• w sqrt(k/m)
• Pendulum (Small oscillations)
• w ?(g/L)

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