Oscillations• 13. Oscillations

• Content

• 13.1 Simple harmonic motion

• 13.2 Energy in simple harmonic motion

• 13.3 Damped and forced oscillations: resonance

• Learning Outcomes

• (a) describe simple examples of free oscillations.

• * (b) investigate the motion of an oscillator using experimental and graphical methods.

• (c) understand and use the terms amplitude, period, frequency, angular frequency and phase difference and express the period in terms of both frequency and angular frequency.

• (d) recognise and use the equation a = – ω2x as the defining equation of simple harmonic motion.

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• (e) recall and use x = xo sin ωt as a solution to the equation a = –ω2x.

• (f) recognise and use v = vo cos ωt, v = ± ω√ (x2o − 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 energy during simple harmonic motion.

• * (i) describe practical examples of damped oscillations with particular reference to the effects of the degree of damping and the importance of critical 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 frequency response and sharpness of the resonance.

• (l) show an appreciation that there are some circumstances in which resonance is useful and other circumstances in which resonance should be avoided. 2

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 atable, i.e. sound waves cause vibrations

One complete movement from the starting point or rest point or equilibriumposition 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 period T 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 theequilibrium position

The amplitude (a scalar quantity) is the maximum displacement

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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.4

Recap from study of waves

Some oscillations maintain a constant period

even when the amplitude of the oscillation

changes. This is known as isochronous and

has been made use of in timing devices

Galelli Galileo discovered this for a

pendulum. A pendulum swinging with 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 oscillation is approximately a sine or

cosine curve

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Simple harmonic motion (shm) A sinusoidal displacement time graph is a

characteristic of an important type of oscillationcalled simple harmonic motion(shm)

In harmonic oscillators the amplitude is constantwith time

SHM is defined as the motion of a particle abouta fixed point such that its force ‘F’ oracceleration ‘a’ is proportional to itsdisplacement ‘x’ from the fixed point, and isdirected towards the point

F is known as the restoring force

Mathematically it is defined as a = - ω2x where ωis the angular frequency and is equal to 2πf

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cont… The defining equation is represented in a graph of a against x as a

straight line of negative 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

Comparisons

In linear motion, acceleration is

constant in magnitude and direction

In circular motion acceleration is

constant in magnitude but not

direction

In simple harmonic motion the

acceleration changes periodically in

magnitude and direction9

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 = - ω2x

which requires mathematics beyond the requirements of

A/AS

• However we need to know the form of the solution

x = x0 sin ωt or x = x0 cos ωt

where x0 is the amplitude (maximum displacement) of

the oscillation

• The solution x = x0 sin ωt is used when at time t = 0, the

particle is at its equilibrium position where x = 0, and

conversely if at time t = 0 the particle is at its maximum

displacement, x = x0 the solution is x = x0 cos ωt

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Velocity & acceleration for shm

• The velocity v of the particle is given by the expressions

v = x0ω cos ωt when x = x0 sin ωt

v = -x0ω sin ωt when x = x0 cos ωt

• The maximum speed is given by v0 = x0ω

• An alternate expression for the velocity is v = ±ω√(x02 –

x2)

(which will be derived next)

• The acceleration a of the particle is given by the

expressions

a = -x0ω2 sin ωt when x = x0 sin ωt

a = -x0ω2 cos ωt when x = x0 cos ωt

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Displacement, velocity and acceleration graphs

x

v

a

t

t

t

Displacement (x), velocity (v) & acceleration time graph

Alternate expression for velocity

• Recall that x = x0 sin ωt and v = x0ω cos ωt

• So sin ωt = x/x0 and cos ωt = v/(x0ω)

• Trigonometric relationship between sine and cosine is

sin2θ + cos2θ = 1

• Applying the above relationship, we have

x2/x02 + v2/(x0

2ω2) = 1 which gives

v2 = x02ω2 - x2 ω2 , hence

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v = ± ω√(x02 - x2)

Example

The displacement x at 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 cos ωt

The amplitude x0 = 0.25 m, ω = 2πf = 7.5 rad/s, therefore f = 1.2 Hz

and period T = 1/f = 0.84 s

b) Substitute t = 0.50 s in the equation

ωt = 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 spring – Hooke’s law

• Consider a mass m suspended from a spring

• The weight mg is balanced by the tension T in the spring

• When the spring is extended downwards by an amount x away from

the equilibrium position, there is an additional upward force called

the restoring force in the spring given by F = - kx

• When the mass is released the restoring force F pulls the mass

upwards 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 oscillation T = 2π√(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 real

experiments we use a finite pendulum bob of finite mass

• When the bob is pulled aside through an angle and released, there will be a restoring

force 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 mg and the tension T in the string

• The component of the weight along the direction of the string mg cos θ, is equal to the

tension T in the string

• The component of the weight at right angles to the direction of the string, mg sin θ , is

the restoring force F. This makes the bob accelerate towards the equilibrium position

• The restoring force depends on θ . As θ increases the restoring force is not proportional

to the displacement and so the motion is oscillatory but not shm, but if the angle is kept

small (less than 5°), θ is proportional to sin θ and exhibits shm (check using your calc)

• The full theory shows period of oscillation T = 2π√(l/g) where g is the acceleration of

free fall

• A simple pendulum can be used to experimentally determine g by repeating the

experiment with different lengths of pendulum and plotting a graph of T2 against 4π2/g

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Hooke’s 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

amplitude oscillations may cause the spring to

become slack

iii) the spring should have no mass; if the mass is > 20x

the mass 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 which gives 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 mis attached to its free end.

a) State 2 conditions that must be met before the subsequent motion may be considered 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 Hooke’s law limit of the spring

b) the mass is given a small vertical displacement such that the spring’s Hooke’s 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)xa

As a is proportional to - x , so resulting motion is shm

i.e. a = - ω2x

a = - (k/m)x = - ω2x, so angular frequency ω = √(k/m)

Therefore period T = 2π/ω = 2π√(m/k)19

Example A light string of length l hangs vertically from a fixed support and a mass m is attached to its

free 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

θ

a) A required condition is that the angular displacement θ 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 θ = ma, so - g sin θ = a

For small θ , sin θ ≈ x/l, so - gx/l ≈ a x

As a is proportional to - x , so resulting motion is shm

i.e. a = - ω2x

Hence, - (g/l)x = - ω2x, so ω = √(g/l) mg

Therefore period T = 2π/ω = 2π√(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

Solution

a) k = F/x, k = 4.9 Nm-1

b) T = 2π√(m/k) T = 0.63 s

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Energy changes in shm

• A system exhibiting simple harmonic motion would possess aconstant total energy at all points of time

• The total energy normally comprises a portion of potentialenergy and another balanced portion of kinetic energy.

• There is thus a continuous interchange of the two energiesduring oscillations.

• For example, a weighted helical spring has a total energy thatis the sum of the kinetic energy of the moving mass and thestored elastic potential energy of the spring.

• Plotting on the same graph for energy versustime/displacement, the two sinusoidal curves are completelyout of phase.

• It can be proven that the total energy of a weighted spring is ½mω2xo

2 which is a constant.

.22

Energy vs time graph

energy

Energy versus time graph

0

0 T/4 T/2 3T/4 T time

total energyK.E. P.E.

Displacement, velocity and acceleration graphs

x

v

a

t

t

t

Displacement (x), velocity (v) & acceleration time graph

Energy changes in shm

• The kinetic energy of a particle of mass m oscillating with shm is½mv2 and from the earlier derivation v2 = x0

2ω2 - x2ω2

• So k.e Ek at displacement x is ½mω2( x02- x2)

• To find the potential energy Ep we need to find the work done againstthe restoring force;

since F = ma , Fres = - mω2x but average restoring force = ½mω2x

• Hence work done = average restoring force x displacement

= ½mω2x2

• The total energy Etot of the oscillating system is given by

• Etot = Ek + Ep = ½mω2( x02- x2) + ½mω2x2

= ½mω2x02

• 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 = ½mω2( x02- x2) + ½mω2x2 = ½mω2x0

2

a) Etot = ½mω2x02 = 2.7 x 10-4 J

b) Ek = ½mω2( x02- x2) = 2.0 x 10-4 J

Ep = ½mω2x2 = 6.7 x 10-5 J

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Natural frequency & resonance

A particle is said to be undergoing free oscillations whenthe 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 systemoscillates in the absence of any damping forces, and it issaid to oscillating in its natural frequency

In real situations, frictional and other resistive forces causethe oscillator’s energy to be dissipated, and this energy isconverted eventually into heat energy. The oscillations aresaid to be damped

When one object vibrates at the same frequency asanother it is said to be in resonance

The ‘swing’ of a frictionless pendulum is an example of afree oscillation.

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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 forced

oscillations and will vibrate at the frequency of the applied force rather than at the

natural frequency of the system

Whether or not the forcing frequency equals the natural frequency, the oscillations are

said 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 frequency by the

driving frequency

When resonance occurs, the amplitude of the resulting oscillations is a maximum as

maximum energy is transferred from the forcing system

E.g. Barton's pendulum – only the pendulum with the same length as the original will

oscillate with the biggest amplitude

Applications – wind 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

into resonance until the main span broke up 28

Damped oscillations

• A damped oscillation is one where frictional forces present

gradually slow down the oscillation and the amplitude decreases

with time i.e. decreasing energy

• Damped oscillations are divided into under-damped, critically

damped and over-damped oscillations

• An under-damped(lightly damped) oscillation is one where the

amplitude of oscillation or displacement of the system decreases

with time. Example: oscillation of a simple pendulum with the

damping or dissipative force as air resistance

• In a critically damped system, oscillations are reduced to zero in the

shortest possible time. Examples: moving coil ammeter or volt

meter, shock absorber, door closer

• In an over-damped(heavy damping) system, a displacement from its

equilibrium position takes a long time for the displacement to be

reduced to zero. Example: door dampers

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Damped oscillations

Displacement vs Time Graph

x

t

under-damped critically-damped

over-damped

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 lower

frequencies

– 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 the required transmission in a receiver

• The natural frequency of an electrical oscillator depends on the

capacitor and 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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14.6 Production and use of ultrasound(a) explain the principles of the generation and detection of ultrasonic waves using piezo-electric transducers(b) explain the main principles behind the use of ultrasound to obtain diagnostic information about internal structures(c) show an understanding of the meaning of acoustic impedance and its importance to the intensity reflection coefficient at a boundary(d) recall and solve problems by using the equation I = I0e–μx for theattenuation of X-rays and of ultrasound in matter

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Ultrasound

Incident

wave Reflected wave

Transmitted wave

Boundary between media

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Recap – piezo-electric effect• Piezo-electric devices contain a crystal which can expand and compress when external

pressure is varied e.g. quartz

• The crystal’s structure is such that the centre of positive charges coincides with the centreof negative charges when not stressed

• When expanded, both centres will not coincide.

• When compressed, the centres will be in the opposite direction as compared to underexpansion.

• The separation results in a voltage across the crystal surface and this effect is known asthe piezo-electric effect

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Piezo-electric transducer• A piezo-electric device is a sensor that detects differences in pressure

(sound wave)• Variation in pressure will result in an ac voltage• The magnitude of the voltage generated depends on the magnitude of the

pressure on the crystal and the polarity depends on whether the crystal iscompressed or expanded i.e. whether the pressure is greater than or lessthan the ambient pressure

• A transducer is any device that converts energy from one form to another• The piezoelectric transducer converts mechanical energy (vibration) into

electrical energy in the form of ac voltages• It also can convert electrical voltages back to vibration.• Hence it acts as a receiver as well as an emitter.• To detect the voltages, opposite faces of the crystal are coated with a

metal (silver) and electrical connections are made to these metal films andsince the voltages are very small they are amplified

• The crystal and its amplifier may be used as a simple microphone forconverting sound signals into electrical signals

• Ultrasound waves may be generated using a piezo-electric crystal suchas quartz, as it can convert electrical voltages to vibrations

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Ultrasound • When a potential difference is applied between the electrodes of the crystal, an

electric field is set up in the crystal which causes forces to act on the ions

• Quartz has a tetrahedral silicate structure with the oxygen ion negatively chargedand the silicon ion positively charged, and as these ions are not held rigidly inposition, they will be displaced slightly when an electric field is applied

• The positive ions will be attracted to the negative electrode, and the negative ionswill be attracted to the positive electrode and depending on the direction of theelectric field, the crystal will become slightly thinner or thicker

• An ac voltage applied across the electrodes will cause the crystal to vibrate with afrequency equal to that of the applied voltage with a small amplitude

• If the frequency of the applied voltage is equal to the natural frequency of vibrationof the crystal, resonance will occur and the amplitude of vibration will be amaximum

• The dimensions of the crystal can be such that the oscillations are in theultrasound region(> 20 kHz) and this will give rise to ultrasound waves in anymedium surrounding the crystal

• In the medical field the ultrasound frequency is in the megahertz region

• Human range of hearing is from 20 Hz to 20,000 Hz

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The reflection and absorption of ultrasound

• Ultrasound is typical of many types of waves in that , when it is incident on a boundarybetween 2 media, some of the wave power is reflected and some transmitted

• For a wave of incident intensity I, reflected intensity IR and transmitted intensity IT, byconservation of energy,

I = IR + IT

• Although for a beam of constant intensity, the sum of the reflected and transmittedintensities is constant, their relative magnitudes depends not only on the angle ofincidence of the beam on the boundary but also on the media themselves

• Hence the relative magnitudes of IR and IT are quantified by reference to the specificacoustic impedance, Z of each media

• The specific acoustic impedance Z is defined as the product of the density ρ of themedium and the speed c of the wave in the medium i.e. Z = ρc

• For a wave incident normally on a boundary between 2 media having specific acousticimpedances of Z1 and Z2, the ratio of the reflected intensity IR to the incident intensity Iknown as the intensity reflection coefficient symbol α

i.e. intensity reflection coefficient α = IR/I = (Z2-Z1)2/(Z2+Z1)2

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Typical values of specific acoustic impedance and speed of ultrasound

Medium Speed/m s-1 specific acoustic impedance/kg m-2 s-1

Air 330 430

Water 1500 1.5 x 106

Blood 1600 1.6 x 106

Fat 1500 1.4 x 106

Muscle 1600 1.7 x 106

Soft tissue 1600 1.6 x 106

Bone 4100 5.6 – 7.8 x 10639

Example

Calculate the intensity reflection coefficient for a parallel beam of ultrasound incident normally on the boundary between:

(1) air and soft tissue, specific acoustic impedances of 430 kg m-2 s-1 and 1.6 x 106 kg m-2 s-1 respectively

(2) muscle and bone, specific acoustic impedance of 1.7 x 106 kg m-2 s-1

and 6.5 x 106 kg m-2 s-1 respectively

Solution

Using α = IR/I = (Z2-Z1)2/(Z2+Z1)2

(1) α = 0.999 almost 1!

(2) α = 0.3440

Linear absorption coefficient • The intensity reflection coefficient for a boundary between air and soft tissue

from the last example is almost unity which means that when ultrasound isincident on the body, very little ultrasound is transmitted into the body

• In order to overcome this, it is important that there is no air between thetransducer and the soft tissue(skin) and this is achieved by using a water basedjelly whose specific acoustic coefficient is approximately 1.5 x 106 kg m-2 s-1

• Once the ultrasound is within the medium, the intensity of the wave is reducedby absorption of energy as it passes through the medium

• This causes heating and appropriate frequencies of ultrasound are actually usedin physiotherapy to assist in sprains and similar injuries

• For a parallel beam, the absorption is approximately exponential and for a beamof ultrasound that is incident normally on a medium of thickness x, thetransmitted intensity I is related to the incident intensity I0 by the expression

I = I0 e-kx or I = I0 exp(-kx)

where k is a constant depending on the medium known as the linear absorptioncoefficient. The unit is cm-1

• The coefficient k depends not only on the medium but also the frequency of theultrasound 41

Typical values of linear absorption coefficient, k of ultrasound

Medium Linear absorption coefficient/cm-1

Water 0.0002

Bone 0.13

Muscle 0.23

Air 1.2

42

Example A parallel beam of ultrasound is incident on the surface of a muscle andpasses through a thickness of 3.5 cm of the muscle. It is then reflected atthe surface of a bone and returns through the muscle to its surface.

Calculate the fraction of the incident intensity that arrives back at thesurface of the muscle given that the linear absorption coefficient formuscle is 0.23 cm-1 and the fraction reflected at bone-muscle interface is0.34(from the last example)

Solution

The beam passes through a total thickness of 7.0 cm of muscle

For the attenuation in the muscle,

using I = I0 exp(-kx) = I0 exp(-0.23 x 7.0) = 0.20I0

Given that the fraction reflected at the bone-muscle interface i.e. α is 0.34,

therefore the fraction received back at surface = 0.34 x 0.20 = 0.068 = 1/15 43

Obtaining diagnostic information using ultrasound

• The ultrasound transducer is placed on the skin with the waterbased jelly excluding any air between the transducer and the skin

• Short pulses of ultrasound are transmitted into the body wherethey are partly reflected and partly transmitted at the boundariesbetween media in the body

• The reflected pulses or echoes return to the transducer wherethey are detected and converted into voltage pulses which areamplified and processed by electronic circuits to be displayed ona screen or oscilloscope

• The time between the transmitted and reflected pulses givesinformation as to the distance of the boundary from thetransducer

• The intensity of the reflected beam gives information as to thenature of the boundary

• 2 techniques are in common use for the display of an ultrasoundscan:– A-scan– B-scan

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A-scan • A short pulse of ultrasound is transmitted into the body through the coupling

medium

• At each boundary some of the energy of the pulse is transmitted and some isreflected

• The transducer(generator/detector) detects the reflected pulses as it now acts as areceiver

• The signal is amplified and displayed on a c.r.o.

• Reflected pulses received at the transducer from deeper in the body tend to havelower intensity than those reflected from boundaries near the skin

• This is due not only to the energy being absorbed by the various media but also onthe return of the reflected pulse to the transducer, some of the energy of the pulsewill again be reflected at intervening boundaries

• To allow for this, echoes received later at the transducer are amplified more thanthose received earlier

• A vertical line is observed on the c.r.o. corresponding to the detection of eachreflected pulse

• The time-base of the c.r.o. is calibrated so that knowing the speed of the ultrasoundwave in each medium, the distance between boundaries can be determined 45

B-scan • This consists of a series of A-scans all taken from

different angles so that a 2-D image can be formed

• The ultrasound probe consisting of agenerator/detector, for a B-scan does not consist of asingle crystal, but rather an array of smaller crystalseach one at a different angle to its neighbours

• The separate signals received from each of the crystalsin the probe is processed and a pattern of spots is builtup to create a 2-D image for immediate viewing,photographing or to be stored in computer memory

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Advantages of ultrasound scanning • Health risk to patient and operator is very much less

compared to use of X-rays

• The equipment is much more portable and easy to use

• Using higher frequency ultrasound enables greaterresolution to be obtained i.e. greater details to be seen

• Modern techniques allow for the detection of very lowintensity reflected pulses, hence boundaries betweentissues where there is little change in acousticimpedance can be detected

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