Post on 16-Jan-2020
PH 221-1D Spring 2013
Oscillations
Lectures 35-37
Chapter 15 (Halliday/Resnick/Walker, Fundamentals of Physics 9th edition)
1
Chapter 15Oscillations
In this chapter we will cover the following topics:
Displacement, velocity and acceleration of a simple harmonic oscillator
Energy of a simple harmonic oscillator
Examples of simple harmonic oscillators: spring-mass system, simple pendulum, physical pendulum, torsion pendulum
Damped harmonic oscillator
Forced oscillations/Resonance2
OscillationsPeriodic motion – the motion of a particle or a system of particles is periodic, or cyclic, if it repeats again and again at regular intervals of time.
Example:• The orbital motion of a planet• The uniform rotational motion of a phonograph turntable• Back and forth motion of a piston in an automobile engine• Vibrations of a guitar string
Oscillation – back and forth or swinging periodic motion is called an oscillation
Simple Harmonic Motion
• Simple harmonic motion is a special kind of one dimensional periodic motion• The particle moves back and forth along a straight line, repeating the same motion again and again
Simple harmonic motion – the particles position can be expressed as a cosine or a sine function of time.
Cosines and sines are called harmonic functions => we call motion of the particle harmonic.
3
When an object attached to a horizontal spring is moved from its equilibrium position and released, the restoring force F = -kx leads to simple harmonic motion
Moving strip of paper at a steady rate
Record position of the vibrating object vs. time
pen
A = amplitude – the maximumdisplacement from equilibrium
x = 0 is the equilibrium position of the object.
Position as a function of time has the shape of trigonometric sine or cosine function
4
In fig.a we show snapshots of a sSimple Harmon
impleoscillat
ic
ing
Motio
syst
n (
em
SHM)
.
The motion is periodic i.e. it repeats in time. The time needed to completeone repetition is known as the The number of
repetitions per u
(symbol , units: s ).
nit time is called the f
perio
n
d
reque
T
cy
The displacement of the particle is given by the equation: Fig.b is a plot of ( ) versu amplitudes . The quantity is called the of the
(symbol
mot
, unit hertz, Hz) 1
( ) c
s
i
o
m
m
fT
x t x tx t
f
t x
o
n. It gives the maximum possible displacement of the oscillating objectThe quantity is called the of the oscillator. It is given by angular frequencythe equation:
( ) cosmx t x t
22 fT 5
( ) cosmx t x t The quantity is called the phase angle of the oscillator.The value of is determined from the displacement (0) and the velocity (0) at 0. In fig.a ( ) is plotted versus for 0. ( )
xv t x t
t x t x
Velocity of SHM
velocity
cos
( )( ) cos sin
The quantity is called the It expresses the maximum possible value of ( ) In fig.b the
am
velocity ( ) is plo
pl
tt
itude
m
m
m m
m
v
t
dx t dv t x t x tdt dt
xv t
v t
ed versus for 0. ( ) sinm
tv t x t
2 2
2
( )( ) sin cos
The quantity is called the
Acceleration of SHM:
acceleration ampli .It expresses the maximumpossible value of a( ). In fig.c the accele
tr
dat
u e am
m m
m
dv t da t x t x t xdt dt
xt
2
ion a( ) is plotted versus for 0. ( ) cosm
t ta t x t
6
2
2 2
We saw that the acceleration of an object undergoing SHM is:
If we apply Newton's second law we get:
The Force Law for Simple Harmonic Motion
Simple harmonic motion occurs whe
a x
F ma m x m x
The force can be writn the force acting on a
ten as: where is a constant. If we compare the two expressions for F we have:
n object is proportionalto the disaplacement but opposite in sign. F
CCx
2
d
anm C
Consider the motion of a mass attached to a springof srping constant than moves on a frictionless horizontal floor as shown in the figure.
mk
The net force on is given by Hooke's law: . If we compare this equation with the expression we identify the constant C with the sping constant k.We can then calculate the angular fre
F m F kxF Cx
quancy and the period .
and 2 2
T
C k m mTm m C k
km
2 mTk
2 mTC
Problem : When an object of mass m1 is hung on a vertical spring and set into vertical simple harmonic motion, its frequency is 12.0 Hz. When another object of mass m2 is hung on the spring along with m1, the frequency of the motion is 4.00 Hz. Find the ratio m2/m1 of the masses.
8
Three springs with force constants k1 = 10.0 N/m, k2 = 12.5 N/m, and k3 = 15.0 N/m are connected in parallel to a mass of 0.500 kg. The mass is then pulled to the right and released. Find the period of the motion.
9
A mass of 0.300 kg is placed on a vertical spring and the spring stretches by 10.0 cm. It is then pulled down an additional 5.00 cm and then released. Find:
a) K; b) ω; c) frequency; d) T; e) max velocity; f) amax;g) Fmax (max restoring force); h) V for x = 2.00 cm;i) The equations for displacement, velocity and acceleration at any time
10
11
Problem 24
12
The mechanical energy of a SHM is the sum of its potenEnergy in S
tial and kiimple Harmonic Motion
netic energies an
d
. E
U K
2 2 2
2 2 2 2 2 2
2 2 2 2
1 1Potential energy cos2 2
1 1 1Kinetic energy sin sin2 2 2
1 1Mechanical energy cos sin2 2
In the figure we plot the potential energy
m
m m
m m
U kx kx t
kK mv m x t m x tm
E U K kx t t kx
U
(green line), the kinetic energy (red line) and the mechanical energy (black line) versus time . While and
vary with time, the energy is a constant. The energy of the oscillating objectt
KE t U
K Eransfers back and forth between potential and kinetic energy, while the sum of
the two remains constant 13
Problem 35. A block of mass M =5.4 kg, at rest on a horizontal frictionless table, is attached to a rigid support by a string of constant k=6000 N/m. A bullet of mass m=9.5 g and velocity
of magnitv yde 630 m/s strikes and is embedded in the block. Assuming the compression of the spring is negligible untill the bullet is embedded, determine (a) the speed of the blockimmidiately after the collison, and (b) the amplitude of the resulting simple harmonic motion
The problem consists of two distinct parts: the completely inelastic collision (which isassumed to occur instantaneously, the bullet embedding itself in the block before theblock moves through significant distance) followed by simple harmonic motion (of massm + M attached to a spring of spring constant k). (a) Momentum conservation readily yields v´ = mv/(m + M). With m = 9.5 g, M = 5.4 kg and v = 630 m/s, we obtain ' 1.1 m/s.v (b) Since v´ occurs at the equilibrium position, then v´ = vm for the simple harmonic motion. The relation vm = xm can be used to solve for xm, or we can pursue the alternate(though related) approach of energy conservation. Here we choose the latter:
2 22 2 2
21 1 1 1'2 2 2 2m m
m vm M v kx m M kxm M
which simplifies to
32
3
(9.5 10 kg)(630 m/s) 3.3 10 m.(6000 N/m)(9.5 10 kg 5.4kg)
mmvx
k m M
14
In the figure we show another type of oscillating systemIt consists of a disc of rotational
An Angular
inertia
Simple Harmonic Oscillator; Tors
suspended from a wire that tw
ion Pendul
ists as t
um
m roI
ates by an angle . The wire exertson the disc a This is the angular form of Hooke's law. The constant
is called th
restoring torque
torsion constane of
the wire t
.
If we compare the expression for the torque with the force equation we realize that we identify the constant C with the torsion constant .
We can thus readily determine the angular frequenF Cx
cy and the period of the
oscillation.
We note that I is the rotational inertia of the disc about an axis that coincides wit
2 2
h the wire. The angle is given b
C I ITI I C
T
y the equation:
( ) cosmt t 15
A simple pendulum consists of a particle of mass suspended by a string of length from a pivot point. If the mass is disturbed from its equilibrium position
Th
th
e Simple P
e net forc
endu
e ac
lu
n
m
ti g
mL
on it is such that the system exectutes simple harmonic motion.There are two forces acting on : The gravitational force and the tension from the string. The net torque of these forces is:
g
m
r F L sin Here is the angle that the thread
makes with the vertical. If 1 (say less than 5 ) then we canmake the following approximation: where is expressed in r
sadians. Wit
in h this
mg
approximation the torque is:
If we compare the expression for Lmg
with the force equation we realize that we identify the constant C with the term . We can thus readily determine the angular frequency
and the period of the oscillation. ; 2
F
C mgL TI
CxLmg
TI
2I IC mgL
In the small angle approximation we assumed that << 1 and used the approximation: sin We are now going to decide what is a “small” angle i.e. up to what angle is the approximation reasonably accurate?
(degrees) (radians) sin5 0.087 0.087
10 0.174 0.174
15 0.262 0.259 (1% off)
20 0.349 0.342 (2% off)
Conclusion: If we keep < 10 ° we make less that 1 % error
17
2
2
Physical Pendul
The rotational inertia about the pivot point is equal to
Thus 2 2
A physical pendulum is an extened rigid body that is suspendedfrom a fixed point O and oscill
um
ates
I mL
I mLTmgL mgL
under the influence of gravityThe net torque sin Here is the distance between point O and the center of mass C of the suspended body.If we make the small angle approximation 1, we have:
mgh h
. If we compare the torques with the
force equation we realize that we identify the constant C with the term . We can thus readily determine period of the oscillatio
2 2
n.
mghF C
I ITC
xhmg
T
Here is the rotational
inertia about an axis through O. h
Img
2 LTg
2 ITmgh
18
Example 1: The pendulum can be used as a very simple device to measure the acceleration of gravity at a particular location.
• measure the length “l” of the pendulum and then set the pendulum into motion• measure “T” by a clock•
2 2 2 2
2 2
2
2
1 1 1( )12 2 3
For a uniform rod o
1/ 3 22 2 2 2 23 ( / 2) 3
83
f length L
comI I mh mL m L mL
I I mL L LTC mgh mgh g L g
LgT
19
1. From the piece of rope and a bob we make a simple pendulum2. We set pendulum into motion3. We measure period “T” by a clock4. We calculate the length of the pendulum (rope)
5. With a help of the rope of the known length we measure the dimensions of the room a x b x h and its volume V = a x b x h
Example : The Length of a Pendulum. A student is in an empty room. He has a piece of rope, a small bob, and a clock. Find the volume of the room.
20
Pr oblem 44. A physical pendulum consists of a meter stick that is pivoted at a small hole drilled through the stick a distance d from the 50 cm mark. The period of oscillation is 2.5 s. Find d?
We use Eq. 15-29 and the parallel-axis theorem I = Icm + mh2 where h = d, the unknown. For a meter stick of mass m, the rotational inertia about its center of mass is Icm = mL2/12 where L = 1.0 m. Thus, for T = 2.5 s, we obtain
T mL mdmgd
Lgd
dg
2 12 212
2 2 2
/ .
Squaring both sides and solving for d leads to the quadratic formula:
2 42 2/2 /2 /3= .
2g T g T L
d
Choosing the plus sign leads to an impossible value for d (d = 1.5 L). If we choose the minus sign, we obtain a physically meaningful result: d = 0.056 m.
2 ITmgd
21
Consider an object moving on a circular path of radius with a uniform speed . If we project th
Simple
e posit
Harmonic Motion and Unifor
ion of themoving particle
m Circular
at point
Moti
P on the
on
mxv
x-axis we get point P.
The coordinate of P is: ( ) cos . While point P executes unifrom circular motion its projection P moves along the x-axis with simple harmonic motion.The speed v of point
mx t x t
P is equal to . The direction of the velocity vector is along the tangentto the circular path. If we project the velocity on the x-axis we get: ( ) sin The acceleration points al
m
m
x
vv t x t
a
2
ong the center O. If we project along the x-axis we get: a( ) cos
Conclusion: Whether we look at the displacement ,the velocity, or the acceleration , the projection of uniform circular m
ma t x t
otion on the x-axis is SHM 22
When the amplitude of an oscillating object is reduced due to the presence of an external force the motion
Damped Simple Harmonic Motion
dampis said to be
An example is given in the figure. A med. ass m attached to a spring of spring constant k oscillates vertically.The oscillating mass is attached to a vane submerged in a
liquid. The liquid exerts a whose magnitu
damping force de is given by t
dF
he equation: dF bv
The negative sign indicates that opposes the motion of the oscillating mass.The parmeter is called the . The net force on m is:
damping From Newton's second
consta la nt
w we
d
net
Fb
F kx bv
2
2
2
2
have:
We substitute with and a with and get
the following differential equation: 0 d x dxm b kxd
dx d xkx bv ma vdt d
t t
t
d
23
2
2
Newton's second law for the damped harmonic oscillator:
0 The solution has the form:d x dxm b kxdt dt
/ 2
In the picture above we plot ( ) versus . We can regard the above solutionas a cosine function with a time-dependent amplitude . The angularfrequency of the damped harmonic oscillator
bt mm
x t tx e
2
22
is given by the equation:
1 For an undamped harmonic oscillator the energy 2
If the oscillator is damped its energy is not constant but decreases with time. If the
k=m
damping is s
4 mE kxbm
/ 2
2 /
mall we can replace with By doing so we find that:1( ) The mechanical energy decreases exponentially with time 2
bt mm m
bt mm
x x e
E t kx e
/ 2 ( ) cosbt mmx t x e t
24
25
Problem 59. The amplitude of a lightly damped oscillator decreases by 3.0% during each cycle. What percentage of the mechanical energy of the oscillator is lost in each cycle?
Since the energy is proportional to the amplitude squared (see Eq. 15-21), we find the fractional change (assumed small) is
= = 2 = 2 . 2
2 2
E E
EdEE
dxx
x dxx
dxx
m
m
m m
m
m
m
Thus, if we approximate the fractional change in xm as dxm/xm, then the above calculation shows that multiplying this by 2 should give the fractional energy change. Therefore, ifxm decreases by 3%, then E must decrease by 6.0 %.
212 mE U K kx
Moving support
If an oscillating system is disturbed and then allowedto oscillate freely the corresponding angular frequency is calle
Forced Oscillations and Resonance
nad the The satural f me systrequ em ce ancy l. n aso
be driven as shown in the figure by a moving support thatoscillates at an arbitrary angular frequency . Such a forced oscillator oscillates at the angular frequency of the driving force. The
d
d
displacement is given by:
The oscillation amplitude varieswith the driving frequency as shown in the lower figure.The amplitude is approximatetly greatest when
( ) co
This condition
s m
d
mx t x xt
is called All mechanical structureshave one or more natural frequencies and if a structure issubjected to a strong external driving force whose frequencymatches one of the natura
resona
l fre
nce.
quencies, the resulting oscillationsmay damage the structure 26
Problem 63. A 100 kg car carrying four 82 kg people travels over a "washboard" dirt roadwith corrugations 4.0 m apart. The car bounces with maximum amplitude when its speed is 16 km/h. When the car stops, and the people get out, how much does the car body rise on its suspension?
With M = 1000 kg and m = 82 kg, we adapt Eq. 15-12 to this situation by writing
24
kT M m
.
If d = 4.0 m is the distance traveled (at constant car speed v) between impulses, then we may write T = d/v, in which case the above equation may be solved for the spring constant:
22 2= 4 .
4v k vk M m
d M m d
Before the people got out, the equilibrium compression is xi = (M + 4m)g/k, and afterward it is xf = Mg/k. Therefore, with v = 16000/3600 = 4.44 m/s, we find the rise ofthe car body on its suspension is
= 4 = 4+ 4 2
= 0.050 . 2
x x mgk
mgM m
dvi f F
HIK m
km
27
28
We wish to find the effective spring constant for the combination of springs. We do this by finding the magnitude F of the force exerted on the mass when the total elongation ofthe springs is x. Then keff = F/x. Suppose the left-hand spring is elongated by x and the right-hand spring is elongated by xr. The left-hand spring exerts a force ofmagnitude k x on the right-hand spring and the right-hand spring exerts a force ofmagnitude kxr on the left-hand spring. By Newton’s third law these must be equal, so x xr . The two elongations must be the same and the total elongation is twice theelongation of either spring: x x 2 . The left-hand spring exerts a force on the block and its magnitude is F k x . Thus k k x x kreff / /2 2 . The block behaves as if it were subject to the force of a single spring, with spring constant k/2. To find the frequency of its motion replace keff in f k m 1 2/ /a f eff with k/2 to obtain
= 12 2
f km
.
With m = 0.245 kg and k = 6430 N/m, the frequency is f = 18.2 Hz.
Problem 26. Two springs are joined in series and connected to a block of mass 0.245 kgthat is set oscillating over a frictionless floor. The springs each have spring constant k=6430 N/m. What is the frequency of oscillations?