oscillations notes
DESCRIPTION
oscillation notes for students of smmeTRANSCRIPT
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WAVES
&
OSCILLATIONS
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WAVES ARE EVERYWHERE!
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Sound Waves result from periodic oscillations of air molecules, which collide with their neighbours and create a disturbance which moves at the speed of sound.
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Electric and Magnetic fields, when oscillated, can create waves which carry energy. At the right frequency, we see electromagnetic waves as Light.
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PERIODIC MOTION IS
EVERYWHERE
Examples of periodic motion
Earth around the sun
Elastic ball bouncing up an down
Quartz crystal in your watch,
computer clock, iPod clock, etc.
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PERIODIC MOTION IS EVERYWHERE
Examples of periodic motion
Heart beat
In taking your pulse, you count 70.0
heartbeats in 1 min.
What is the period, in seconds, of your
heart's oscillations?
Period is the time for one
oscillation
T= 60 sec/ 70.0 = 0.86 s
What is the frequency?
f = 1 / T = 1.17 Hz
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SOME OSCILLATIONS ARE NOT
SINUSOIDAL:
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A SPECIAL KIND OF PERIODIC OSCILLATOR:
HARMONIC OSCILLATOR
WHAT DO ALL “HARMONIC OSCILLATORS”
HAVE IN COMMON?
1. A position of equilibrium
2. A restoring force, (which may be linear )
Hooke’s law spring F = -k x
In a pendulum the behavior only linear for
small angles:
F=-mg sinθ≈-mg θ where θ = s / L. In this
limit we have: F = -ks with k = mg/L
3. Inertia
4. The drag forces are reasonably small
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SIMPLE HARMONIC MOTION (SHM)
Pull block to the right until x = A
After the block is released from x = A
it will move to the left until it reaches
equilibrium, continue moving to the
left due to inertia until it reaches x=-A
and stop there. Due to restoring
force moves to the right
k m
k m
k m
-A A 0(≡Xeq)
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SIMPLE HARMONIC MOTION (SHM)
The time it takes the block to complete one cycle is called the
period.
Usually, the period is denoted by T and is measured in
seconds.
The frequency, denoted f, is the number of cycles that are
completed per unit of time: f = 1 / T.
In SI units, f is measured in inverse seconds, or hertz (Hz).
If the period is doubled, the frequency is
A. unchanged
B. doubled
C. halved
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SHM DYNAMICS: NEWTON’S LAWS STILL
APPLY
At any given instant we know that F = ma must be true.
But in this case F = -k x and
So: -
k
x
m
F = -k x
a
xm
k
dt
xd
2
2
a differential equation for x(t) !
2
2
dt
xdmmakx
2
2
dt
xdmma
“Simple approach, guess a solution and see if it works!
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SHM SOLUTION...
Try either cos ( t ) or sin ( t )
Below is a drawing of A cos ( t )
where A = amplitude of oscillation
[with = (k/m)½ and = 2 f = 2 /T ]
Both sin and cosine work so need to include
both
T = 2/
A
A
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COMBINING SIN AND COSINE
SOLUTIONS
k
x
m
0
Use “initial conditions” to determine phase !
sin cos
x(t) = B cos t + C sin t
= A cos ( t + )
= A (cos t cos – sin t sin )
=A cos cos t – A sin sin t
Notice that B = A cos C = -A sin tan = -C/B
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WORK “INTEGRAL” Since work is force times distance, and the force
ramps up as we compress the spring further…
takes more work (area of rectangle) to compress a little
bit more (width of rectangle) as force increases (height of
rectangle)
if full distance compressed is kx, then force is kx, and
area under force “curve” is ½(base)(height) = ½(x)kx =
½kx2
area under curve is called an integral: work is integral of
force
F
orc
e
F
orc
e
F
orc
e
distance (x) distance (x) distance (x)
Area is a work: a
force (height) times
a distance (width) Force from spring
increases as it is
compressed
further
Total work done is
area of triangle
under force
curve
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ENERGY OF THE SPRING-MASS
SYSTEM
Kinetic energy is always
K = ½ mv2 = ½ m(A)2 sin2(t+)
Potential energy of a spring is,
U = ½ k x2 = ½ k A2 cos2(t + )
And 2 = k / m or k = m 2
U = ½ m 2 A2 cos2(t + )
K + U = ½ m 2 A2 = ½ k A2 which is constant
x(t) = A cos( t + )
v(t) = dx/dt = -A sin( t + )
a(t) = dv/dt = -2A cos( t + )
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This diagram shows how the energy transforms from potential to kinetic and back, while the total energy remains the same.
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EXAMPLE:
USING CONSERVATION OF
ENERGY
M
arc
h 1
5, 2
01
3
Ph
ysic
s 1
14
A - L
ectu
re 3
2
A 500 g block on a spring is pulled a distance of 20 cm and released. The subsequent oscillations are measured to have a period of 0.80 s. At what position (or positions) is the speed of the block 1.0 m/s?
2 2 2 2kv Ax Ax
m
22
2 2(1.0 m/s)(0.20 m) 0.154 m15.4 cm(7.85 rad/s)
vxA
220.80 s so 7.85 rad/s
(0.80 s)T
T
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QUESTION
P
hysic
s 1
14
A - L
ectu
re 3
2
Four springs have been compressed from their equilibrium positions at x = 0.
Which system has the largest maximum speed in its oscillation?
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The Simple Pendulum A simple pendulum consists of a mass m (of negligible size) suspended by a string or rod of length L (and negligible mass).
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M
arc
h 1
5, 2
01
3
Ph
ysic
s 1
14
A - L
ectu
re 3
2
24/32
The Simple Pendulum
Looking at the forces on the pendulum bob, we see that the restoring force is proportional to sin, whereas the restoring force for a spring is proportional to the displacement (which is in this case).
sinF mg mg
This is not a Hooke’s Law force.
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The Small Angle Approximation
However, for small angles, sin θ and θ are approximately equal.
3 5
sin3! 5!
if 1
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Fy = may = T – mg cos()
Fx = max = -mg sin()
If small then x L and sin() dx/dt = L d/dt
ax = d2x/dt2 = L d2/dt2
ax = -g = L d2/ dt2
L d2/ dt2 + g = 0
The solution of the above equation is
= cos(t + ) or = sin(t + )
with = (g/L)½
g
LT
2
2
Note that T does not depend on m, the mass of the pendulum bob.
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A TORSION OSCILLATOR
Ph
ysic
s 1
14
A - L
ectu
re 3
2
The restoring torque is proportional to the angular displacement
The minus sign shows that the torque is directed opposite to the angular displacement
2
2
dt
dII
where I is the rotational inertia
Idt
d
dt
dI
2
2
2
2
This equation is similar to the equation for linear SHM i.e., xImk , ,
)cos( tm
where ω is angular frequency. Similarly
IT 2
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P
hysic
s 1
14
A - L
ectu
re 3
2
The Physical Pendulum
A physical pendulum is an extended solid mass that oscillates around its center of mass.
It cannot be modeled as a point mass suspended by a massless string because there is energy in the rotation of the object as its center of mass moves.
Examples:
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P
In figure, a body of irregular shape is pivoted at P and is
displaced from the equilibrium position by an angle θ . The
centre of mass C lies vertically below P. The restoring torque
is given as
sinmgd
For small θ
mgd
The time period will be
2
2
4
as 2
mgdTI
mgdmgd
IT
So we can find moment of inertia of irregular object by this formula.
Simple pendulum is a special case of physical pendulum.
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EXAMPLE: A COMFORTABLE
PACE
Ph
ysic
s 1
14
A - L
ectu
re 3
2
The pace of a comfortable walk can be estimated if we model each leg as a swinging physical pendulum.
1 12
rod3 2, , and 0.9 mI ML L L
12
rod 3
2 2
/2 /2 22 2 2
3/2 /2
MLIL L LT
g g gML ML
2
2 2(0.9 m)(10 steps)51010 7.7 s
3 3(9.81 m/s)
Lt T
g
Test: 10 steps takes 6.7 s7.7 s
Note that pace period is proportional to L½.
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EXAMPLE: A BLOCK ON A
SPRING A 2.00 kg block is attached to a spring as shown. The force constant of the spring is k = 196 N/m. The block is held a distance of 5.00 cm from equilibrium and released at t = 0.
(a) Find the angular frequency , the frequency f, and the period T. (b) Write an equation for x vs. time.
(196 N/m)9.90 rad/s
(2.00 kg)
k
m
(9.90 rad/s)1.58 Hz
2 2f
1/ 0.635 sT f 5.00 cm and 0A
(5.00 cm)cos(9.90 rad/s)x t
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EXAMPLE: A SYSTEM IN SHM
M
arc
h 1
4, 2
01
3
Ph
ysic
s 1
14
A - L
ectu
re 3
1
A mass attached to a spring, pulled 20 cm to the right, and released at t=0. It makes 15 complete oscillations in 10 s.
a. What is the period of oscillation?
b. What is the object’s maximum speed?
c. What is its position and velocity at t=0.80 s?
15 oscillations
10 s
1.5 oscillations/s1.5 Hz
f
1/ 0.667 sT f
max
2 2(0.20 m)1.88 m/s
(0.667 s)
Av
T
2 2(0.80 s)cos(0.20 m)cos 0.062 m6.2 cm
(0.667 s)
txAT
max
2 2(0.80 s)sin(1.88 m/s)sin 1.79 m/s
(0.667 s)
tvv
T
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EXAMPLE: FINDING THE TIME
A mass, oscillating in simple harmonic motion, starts at x = A and has period T.
At what time, as a fraction of T, does the mass first pass through x = ½A?
111
62cos
2 23
T Tt T
1
2
2cos
tx AA
T
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If the springs in your 1000 kg car compress by 10
cm (e.g., when lowered off of jacks):
then the springs must be exerting mg = 10,000 Newtons
of force to support the car
F = kx = 10,000 N, x = 0.1 m
so k = 100,000 N/m (stiff spring)
this is the collective spring constant: they all add to this
Now if you pile 400 kg into your car, how much will
it sink?
4,000 = (100,000)x, so x = 4/100 = 0.04 m = 4 cm
Could have taken short-cut:
springs are linear, so 400 additional kg will depress car
an additional 40% (400/1000) of its initial depression
EXAMPLE