section iv 14 oscillations
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Oscillations
14. Oscillations Content
14.1 Simple harmonic motion
14.2 Energy in simple harmonic motion
14.3 Damped and forced oscillations: resonance
Learning Outcomes (a) describe simple examples of free oscillations.
* (b) investigate the motion of an oscillator using experimentaland graphical methods.
(c) understand and use the terms amplitude, period, frequency,
angular frequency and phase difference and express theperiod in terms of both frequency and angular frequency.
(d) recognise and use the equation a =2x as the definingequation of simple harmonic motion.
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(e) recall and use x = xo sin t as a solution to the equation a = 2
x. (f) recognise and use v = vocos t, v = (x2
o x2)
* (g) describe with graphical illustrations, the changes in displacement,velocity and acceleration during simple harmonic motion.
(h) describe the interchange between kinetic and potential energyduring simple harmonic motion.
* (i) describe practical examples of damped oscillations with particular
reference to the effects of the degree of damping and the importance ofcritical damping in cases such as a car suspension system. (j) describe practical examples of forced oscillations and resonance. * (k) describe graphically how the amplitude of a forced oscillation
changes with frequency near to the natural frequency of the system,and understand qualitatively the factors which determine the frequencyresponse and sharpness of the resonance.
(l) show an appreciation that there are some circumstances in which
resonance is useful and other circumstances in which resonanceshould be avoided.
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Oscillations and vibrations
Vibrations and oscillations occur all the time and are everywhere.
Vibrations are physical evidence of waves, such as a loud stereo shaking a
table, i.e. sound waves cause vibrations
One complete movement from the starting point or rest point or equilibrium
position and back to the starting point or rest position or equilibrium positionis known as an oscillation
The time taken for one complete oscillation is referred to as the periodT ofthe oscillation
The number of oscillations per unit time is the frequency f
Frequency f = 1/T , may be measured in hertz (1 Hertz = 1 s-1) or in min-1,hour-1 etc
The distance from the equilibrium position is known as the displacement andit is a vector quantity since the displacement may be on either side of the
equilibrium position
The amplitude(a scalar quantity) is the maximum displacement 3
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Examples of oscillatory motion
beating of a heart
a simple pendulum
a vibrating guitar string
vibrating tuning fork
atoms in solids
air molecules oscillate when sound waves travel through air.
oscillations in electromagnetic waves such as light and radio waves
oscillations in alternating current and voltage.
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Recap from study of waves
Some oscillations maintain a constant period even when the amplitudeof the oscillation changes. This is known as isochronous and has beenmade use of in timing devices
Galelli Galileo discovered this for a pendulum. A pendulum swingingwith a large amplitude is not isochronous
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Displacement-time graphs
It is possible to plot displacement-time graphs for oscillators
The graph describing the variation of displacement with time may have
different shapes depending on the oscillating system
For many oscillators the displacement-time graph of a free oscillationis approximately a sine or cosine curve
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Simple harmonic motion (shm)
A sinusoidal displacement time graph is a characteristic of animportant type of oscillation called simple harmonic motion(shm)
In harmonic oscillators the amplitude is constant with time
SHM is defined as the motion of a particle about a fixed point suchthat its force F or acceleration a is proportional to its displacement x
from the fixed point, and is directed towards the point F is known as the restoring force
Mathematically it is defined as a = - 2x where is the angularfrequency and is equal to 2f
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cont The defining equation is represented in a graph of a against x as a
straight line ofnegative gradient through the origin.
Gradient is negative because of the minus sign in the equation whichrepresents that acceleration is always directed towards the fixed pointfrom which the displacement is measured
This means that in shm, acceleration is directly proportional to thedisplacement/distance from the fixed point and is always directed tothat point
Acceleration is always opposite to the displacement since the force isalso opposite to the displacement
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a
x0
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Comparisons
In linear motion, acceleration is constant in magnitude and direction
In circular motion acceleration is constant in magnitude but notdirection
In simple harmonic motion the acceleration changes periodically inmagnitude and direction
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Solution of equation for shm
In order to find the displacement time relation for a particle moving in
shm, we need to solve the equation a = - 2x which requiresmathematics beyond the requirements of A/AS
However we need to know the form of the solution
x = x0 sint or x = x0 costwhere x0 is the amplitude of the oscillation
The solution x = x0 sint is used when at time t= 0, the particle is atits equilibrium position wherex = 0, and conversely if at time t= 0 the
particle is at its maximum displacement,x = x0 the solution is x = x0
cost
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Velocity & acceleration for shm
The velocityvof the particle is given by the expressions
v = x0 cost when x = x0 sint
v = -x0 sint when x = x0 cost
The maximum speed is given by v0= x0
An alternate expression for the velocity is v =(x02x2)
(which will be derived next)
The accelerationaof the particle is given by the expressions
a = -x02sint when x = x0 sint
a = -x02cost when x = x0 cost
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Displacement, velocity and acceleration graphs
x
v
a
t
t
t
Displacement (x), velocity (v) & acceleration time graph
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Alternate expression for velocity
Recall that x = x0 sint and v = x0 cost
So sint = x/x0 and cost = v/(x0)
Trigonometric relationship between sine and cosine is
sin2 + cos2 = 1
Applying the above relationship, we have
x2/x02 + v2/(x0
22) = 1 which gives
v2 = x022 - x2 2 , hence
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v = (x02 - x2)
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Example
The displacement xat time t of a particle moving in shm is given by x =0.25 cos 7.5t where x is in metres and t is in seconds.
a) use the equation to find the amplitude, frequency and period for the
motion
b) find the displacement when t = 0.50 s
Solution
a) Compare the equation with x = x0 cost
The amplitude x0 = 0.25 m, = 2f = 7.5 rad/s, therefore f = 1.2 Hzand period T = 1/f = 0.84 s
b) Substitute t = 0.50 s in the equationt = 7.5 x 0.50 = 3.75 rad = 215
so x = 0.25 cos 215 = -0.20 m
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Worked examples of shm
Mass on a helical spring
Simple pendulum
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Mass on a helical springHookes law
Consider a mass msuspended from a spring The weight mg is balanced by the tensionT in the spring
When the spring is extended downwards by an amount x away fromthe equilibrium position, there is an additional upward force called the
restoring forcein the spring given by F = - kx
When the mass is released the restoring force F pulls the massupwards towards the equilibrium position. The minus sign shows the
direction of this force.
As the force is proportional to the displacement, the acceleration is
also proportional to the displacement and is directed towards the
equilibrium position meeting the condition for shm
The full theory shows that the period of oscillationT = 2(m/k)
since F = ma, then ma = - kx
hence a = - (k/m)x = dv/dt = d2x/dt2
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The simple pendulum
A simple pendulum is a point mass m on a light inelastic string although in realexperiments we use a finite pendulum bob of finite mass
When the bob is pulled aside through an angle and released, there will be a restoringforce acting in the direction of the equilibrium position
Because the pendulum moves in an arc of a circle, the displacement will be an angular
displacement rather than a linear displacement
The 2 forces on the bob are its weight mgand the tensionT in the string
The component of the weight along the direction of the string mg cos, is equal to thetensionT in the string
The component of the weight at right angles to the direction of the string, mg sin , isthe restoring force F.This makes the bob accelerate towards the equilibrium position
The restoring force depends on . As increases the restoring force is notproportional to the displacement and so the motion is oscillatory but not shm, but if the
angle is kept small (less than 5), is proportional to sin and exhibits shm (checkusing your calc)
The full theory shows period of oscillationT = 2(l/g) where g is the acceleration offree fall
A simple pendulum can be used to experimentally determine g by repeating the
experiment with different lengths of pendulum and plotting a graph ofT2
against 42
/g 17
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Hookes law & shm
Any system which obeys Hooke's Law exhibits shm
but i) extensions must not exceed the limit of proportionality
ii) the spring must have small oscillations as large amplitudeoscillations may cause the spring to become slack
iii) the spring should have no mass; if the mass is > 20x themass of the spring, the error is 1%
This example of shm is a particularly useful model for interatomic forces
and vibration of molecules containing atoms oscillating as if connected by
tiny springs
The frequency of oscillation can be measured using spectroscopy whichgives direct information about the bonding
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Example
A light spring of spring constant k hangs vertically from a fixed point and a mass misattached to its free end.
a) State 2 conditions that must be met before the subsequent motion may beconsidered to be simple harmonic.
b) Derive an expression for the period T of the resulting motion.
Solution
a) 2 conditions for shm are:
a) The equilibrium position due to the mass is within the Hookes law limit of the spring b) the mass is given a small vertical displacement such that the springs Hookes law
limit is not exceeded
b) Let x = displacement of mass m, a = acceleration of mass m, F = ma = -kx
Force in a spring is, - kx = ma , hence a = - (k/m)x
As a is proportional to - x , so resulting motion is shmi.e. a = - 2x
a = - (k/m)x = - 2x, so angular frequency = (k/m)Therefore period T =2/ =2(m/k)
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ExampleA light string of length l hangs vertically from a fixed support and a mass m is attached to itsfree end. The mass is given a horizontal displacement and released to swing freely.
a) State a condition which must be satisfied before the resulting oscillation may be
considered shm.
b) Derive an expression for the period T of the resulting motion.
Solution
a) A required condition is that the angular displacement is small l
b) Let x = displacement of mass m, a = acceleration of mass m
In the direction perpendicular to string, F = ma
- mg sin = ma, so - g sin = a
For small , sin x/l, so - gx/l a x
As a is proportional to - x , so resulting motion is shm
i.e. a = - 2x
Hence, - (g/l)x = - 2x, so = (g/l) mg
Therefore period T = 2/ = 2(l/g)
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Example
A helical spring is clamped at one end and hangs vertically. It extends by 10
cm when a mass of 50 g is hung from its free end.
Calculate: a) the spring constant of the spring
b) the period of small amplitude oscillations of the mass
Solutiona) k = F/x, k = 4.9 Nm-1
b) T = 2(m/k) T = 0.63 s
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Energy changes in shm
A system exhibiting simple harmonic motion would possess a constant totalenergy at all points of time
The total energy normally comprises a portion of potential energy andanother balanced portion of kinetic energy.
There is thus a continuous interchange of the two energies duringoscillations.
For example, a weighted helical spring has a total energy that is the sum ofthe kinetic energy of the moving mass and the stored elastic potentialenergy of the spring.
Plotting on the same graph for energy versus time/displacement, the twosinusoidal curves are completely out of phase.
It can be proven that the total energy of a weighted spring is m2
xo2
which is a constant.
.
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Energy vs time graph
energy
Energy versus time graph
0
0 T/4 T/2 3T/4 T time
total energyK.E. P.E.
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Displacement, velocity and acceleration graphs
x
v
a
t
t
t
Displacement (x), velocity (v) & acceleration time graph
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Energy changes in shm
The kinetic energy of a particle of mass m oscillating with shm ismv2and from the earlier derivation v2= x022- x22
So k.e Ek at displacement x is m2(x02- x2) To find the potential energy Epwe need to find the work done against
the restoring force;
since F = ma , Fres = - m2x but averagerestoring force = m2x Hence work done = average restoring force x displacement
= m2x2
The total energy Etotof the oscillating system is given by
Etot= Ek+Ep = m2(x02- x2) + m2x2
= m2x02
This total energy is constant as it merely expresses the law ofconservation of energy
Pg 272 Chris Mee figs 10.22, 10.23,10.24
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Example
A particle of mass 60 g oscillates with shm with angular frequency of 6.3
rad/s and amplitude 15 mm.
Calculate a) the total energy
b) the k.e and p.e at half amplitude (i.e. at x = 7.5 mm)
Solution
Etot= Ek+Ep = m2(x02- x2) + m2x2= m2x02
a) Etot= m2x0
2 = 2.7 x 10-4 J
b) Ek= m2(x0
2- x2) = 2.0 x 10-4 J
Ep = m2x2 = 0.7 x 10-4 J
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Natural frequency & resonance
A particle is said to beundergoing free oscillations when the only
external force acting on it is the restoring force
The total energy remains constant at all points of time
A free oscillation is one where an object or system oscillates in theabsence of any damping forces, and it is said to oscillating in itsnatural frequency
In real situations, frictional and other resistive forces cause theoscillators energy to be dissipated, and this energy is convertedeventually into heat energy. The oscillations are said to be damped
When one object vibrates at the same frequency as another it is saidto be in resonance
The swing of a frictionless pendulum is an example of a freeoscillation.
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Resonance
In the absence of external forces to an oscillating system, the system oscillates at its
natural frequency f0. The only forces acting are the internal forces of the oscillating
system
When an external force is applied to an oscillating system, the system is under forcedoscillations and will vibrate at the frequency of the applied force rather than at thenatural frequency of the system
Whether or not the forcing frequency equals the natural frequency, the oscillations aresaid to be forced when a periodic force acts.
When the forcing frequency is equal to the natural frequency, net energy is taken in
and the amplitude of oscillation builds up further and the applied periodic force is said to
have set the system in resonance. Under such condition, further resonance will result in
more energy being taken in to build up the amplitude further.
Resonance occurs when a system is forced to oscillate at its natural frequencyby thedriving frequency
When resonance occurs, the amplitude of the resulting oscillations is a maximum asmaximum energy is transferred from the forcing system
E.g. Barton's pendulumonly the pendulum with the same length as the original willoscillate with the biggest amplitude
Applicationswind instruments, excessive noise from a moving bus, radio & tv tuning
The Tacoma Narrows suspension bridge in Washington State, USA in 1940 collapsed due
to a moderate gale (of same frequency as natural frequency of bridge) setting the bridge
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Damped oscillations
A damped oscillationis one where frictional forces present
gradually slow down the oscillation and the amplitude decreaseswith time i.e. decreasing energy
Damped oscillations are divided into under-damped, criticallydamped and over-damped oscillations
An under-damped(lightly damped) oscillation is one where theamplitude of oscillation or displacement of the system decreaseswith time. Example: oscillation of a simple pendulum with thedamping or dissipative force as air resistance
In a critically dampedsystem, oscillations are reduced to zero in theshortest possible time. Examples: moving coil ammeter or volt
meter, shock absorber, door closer
In an over-damped(heavy damping) system, a displacement from itsequilibrium position takes a long timefor the displacement to bereduced to zero. Example: door dampers
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Damped oscillations
Displacement vs Time Graph
x
t
under-damped critically-damped
over-damped
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Effects of damping on forced oscillations
Pg 277 Chris Mee fig 10.29 & 10.30
As the degree of damping increases:
The amplitude of oscillation at all frequencies is reduced
The frequency at max amplitude shifts gradually towards lowerfrequencies
The peak becomes flatter
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Electrical resonance
Electrical oscillators made from combinations of capacitors and
inductors(coils) can also be forced into oscillations or be made to resonate
This is the basis of tuning in electronic circuits which pick out therequired transmission in a receiver
The natural frequency of an electrical oscillator depends on the capacitorand inductance of the coil used. By varying the capacitance, we can tune in
to different channels
The range of frequencies selected depends on the damping which in turn
depends on the resistance in the circuit
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