Post on 13-Jan-2016
transcript
Oscillations
Mass on a spring
Particle moving in uniform circular motion
Me walking down lecture hall steps
Oscillation or vibration – the motion of an object that regularly repeats itself, back and forth, over the same path.
That motion is periodic, meaning it has a certain frequency and period.
Springs and Simple Harmonic Motion
€
rF S = −k
r x = m
r a
r a = −
k
mr x
€
ETotal =1
2mv 2 +
1
2kx 2
Eat turning point =1
2kA2
1
2kA2 =
1
2mv 2 +
1
2kx 2
mv 2 = kA2 − kx 2
v = ±k
mA2 − x 2
( )
Recall the spring:
Circles and Simple Harmonic MotionA fundamental parallel can be drawn between SHM (mass on a spring), and uniform circular motion.
€
x = Acos θ( )
θ = θ 0 +ωt
x = Acos ωt +θ 0( )
What is ω for a spring?
Circles and Simple Harmonic Motion
€
1
2kA2 =
1
2mvmax
2
vmax = Ak
m
T =2πA
vmax
= 2πm
k
f =1
T=
1
2π
k
m
ω = 2πf =k
m
A fundamental parallel can be drawn between SHM (mass on a spring), and uniform circular motion.
Position Conceptualized
€
x(t) = Acos(ωt) = Acos2πt
T
⎛
⎝ ⎜
⎞
⎠ ⎟
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x(t) = Asin(ωt) = Asin2πt
T
⎛
⎝ ⎜
⎞
⎠ ⎟
Depending upon the initial conditions…
t = 0
t = 0
T
Velocity
€
v =k
mA2 − x 2
( )
=k
mA2 − Acos ωt +θ 0( )( )
2 ⎛ ⎝ ⎜ ⎞
⎠ ⎟
= Ak
m1 − cos ωt +θ 0( )
2
= Ak
msin ωt +θ 0( )
2
= −Ak
msin ωt +θ 0( )
Acceleration
€
a =F
m
=1
m−kx( )
= −k
mA
⎛
⎝ ⎜
⎞
⎠ ⎟cos ωt +θ 0( )
All together now…
€
x(t) = Acos ωt +θ 0( )
v(t) = −Ak
msin ωt +θ 0( )
a(t) = −k
mA
⎛
⎝ ⎜
⎞
⎠ ⎟cos ωt +θ 0( )
€
xmax = A
vmax = Aω
amax = Aω 2
ExampleAs mass-spring system oscillates with an amplitude of 3.0 cm. If the spring constant is 270 N/m and object has a mass of 0.50 kg, determine the period, the maximum speed and maximum acceleration of the mass.
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T = 2π mk = 2π 0.5
270 = 0.270 s
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vmax = Aω = A km = 0.03 270
0.5 = 0.697 ms
€
amax = Aω 2 = Ak
m= 0.03
270
0.5=16.2 m
s2
3.0 cm
The Pendulum
€
FRestoring = −mgsin θ( )
≈ −mgs
l
⎛
⎝ ⎜
⎞
⎠ ⎟
= −mg
l
⎛
⎝ ⎜
⎞
⎠ ⎟s
= −keff x
€
keff ≡mg
lx ≡ s
The Pendulum
€
keff ≡mg
lx ≡ s
€
ω =k
m=
mgl( )
m=
g
l
T =2π
ω= 2π
l
g
ExampleA simple pendulum has a length of 52.3 cm and makes 83.9 complete oscillations in 2.00 min after being pulled to the side by 0.25m.
Find the period of the pendulum, the acceleration due to gravity at the location of the pendulum, and the maximum speed of the pendulum.
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T =time
oscillation=
120
83.9=1.43 s
€
Tpend = 2π lg
g =2π
T
⎛
⎝ ⎜
⎞
⎠ ⎟2
l
=2π
1.43
⎛
⎝ ⎜
⎞
⎠ ⎟2
0.523
=10.1 ms2
€
vmax = A gl
= 0.25 10.10.523
=1.10 msec
Grandfather ClockYour grandfather clock is running slowly. Every 24 hours, you find that it has lost 2 minutes. How will you fix this problem?
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Tdesired = 2.0 s
Tactual =time for n oscillations
n oscillations
n = 24 hr( ) 60 minhr( ) − 2 min( ) 30 osc
min( )
= 43140
Tactual =24 * 3600
43140= 2.00278 s
€
Tpendulum = 2πl
g
l =T
2π
⎛
⎝ ⎜
⎞
⎠ ⎟2
g
lactual =2.00278
2π
⎛
⎝ ⎜
⎞
⎠ ⎟2
9.8( ) = .996 m
ldesired =2.00
2π
⎛
⎝ ⎜
⎞
⎠ ⎟2
9.8( ) = .993 m
Shorten the pendulum by 3 mm.