phys222 – lssu – bazlurslide 1 chapter 10 simple harmonic motion and elasticity

PHYS222 – LSSU – Bazlur Slide 1

Chapter 10

Simple Harmonic Motion and

Elasticity

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Force is proportional to displacementFor small displacements, the force required to stretch or

compress a spring is directly proportional to the displacement x, or

The constant k is called the spring constant or stiffness of the spring.

A spring behaves according to the above equation is said to be an ideal spring.

xF Appliedx

xkF Appliedx

AppliedxF

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10.1 The Ideal Spring and Simple Harmonic Motion

xkF Appliedx

spring constant

Units: N/m

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10.1 The Ideal Spring and Simple Harmonic Motion

Example-1: A Tire Pressure Gauge

The spring constant of the springis 320 N/m and the bar indicatorextends 2.0 cm.

What force does theair in the tire apply to the spring?

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10.1 The Ideal Spring and Simple Harmonic Motion

N 4.6

m 020.0mN320

xkF Appliedx

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Tire Pressure?

Tire pressure varies from 2 bar to 2.2 bar.

Or, 28 psi to 32 psi

1 bar = 100 kPa = 100 000 Pa = 100 000 N/m2

Pressure = Force per unit Area

P = F/A

What is the area of the Tire Pressure Gauge?

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10.1 The Ideal Spring and Simple Harmonic Motion

Conceptual Example-2: Are Shorter Springs Stiffer?

A 10-coil spring has a spring constant k. If the spring iscut in half, so there are two 5-coil springs, what is the springconstant of each of the smaller springs?

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Shorter Springs are StifferThe spring constant of each 5-coil spring is 2k.

Spring constant 1/# of coils

k for 10 coils

2k for 5 coils

10k for 1 coil

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Restoring ForceTo stretch or compress a spring, a force must be applied

to it.

In accord with Newton’s third law, the spring exert an oppositely directed force of equal magnitude.

This reaction force is applied by the spring to the agent that does the pulling or pushing.

The reaction force is also called a “restoring force”.

Fx = - kx

The reaction force always points in a direction opposite to the displacement of the spring from its unstrained length.

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Restoring ForceThe restoring force of a spring can also contribute to the

net external force.

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10.1 The Ideal Spring and Simple Harmonic Motion

HOOKE’S LAW: RESTORING FORCE OF AN IDEAL SPRING

The restoring force on an ideal spring is xkFx

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Why it is called Restoring ForceAn object of mass m is

attached to a spring on a frictionless table.

In part A, the spring has been stretched to the right, so the spring exerts the left-ward pointing force Fx.

When the object is released, this force pulls it to the left, restoring it toward its equilibrium position.

xkF Appliedx

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Restoring Force causes Simple Harmonic Motion

When the object is released, the restoring force pulls it to its equilibrium position.

In accord with Newton’s first law, the moving object has inertia and coasts beyond the equilibrium position, compressing the spring as in part B.

The restoring force exerted by the spring now points to the right and after bringing the object to a momentary halt, acts to restore the object to its equilibrium position.

Since no friction acts on the object the back-and-forth motion repeats itself.

xkF Appliedx

When the restoring force has the mathematical form given by Fx = -kx, the type of friction-free motion is designated as “simple harmonic motion”.

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Simple Harmonic Motion

By attaching a pen to the object and moving a strip of paper past it at a steady rate, we can record the position of the vibrating object as time passes.

The shape of this graph is characteristic of simple harmonic motion and is called sinusoidal.

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Simple Harmonic MotionSimple harmonic motion, like any motion, can be

described in terms of – displacement, – velocity, and – acceleration.

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10.2 Simple Harmonic Motion and the Reference Circle

tAAx coscos

DISPLACEMENT

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10.2 Simple Harmonic Motion and the Reference Circle

tAAx coscos

Radius = A

The displacement of the shadow, x is just the projection of the radius A onto the x-axis:

Where = t

, the angular speed in rad/s

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Angular Speed, = /t rad/s = t rad

Angular displacement () for one cycle is 2 rad in T.

So, = 2/T

= 2Because is directly proportional to the frequency ,

is often called the angular frequency.

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10.2 Simple Harmonic Motion and the Reference Circle

period T: the time required to complete one cycle

frequency f: the number of cycles per second (measured in Hz)

Tf

1 f

T 2

2

amplitude A: the maximum displacement

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10.2 Simple Harmonic Motion and the Reference Circle

VELOCITY

tAvvv

Tx sinsinmax

AwrwvT

Tangential velocity, VT

= Radius x angular velocity

Velocity of the shadow, Vx

= X component of VT

Maximum Velocity of the shadow, Vx = A

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10.2 Simple Harmonic Motion and the Reference Circle

Example 3 The Maximum Speed of a Loudspeaker Diaphragm

The frequency of motion is 1.0 KHz and the amplitude is 0.20 mm. (a)What is the maximum speed of the diaphragm?(b)Where in the motion does this maximum speed occur?

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10.2 Simple Harmonic Motion and the Reference Circle

tAvvv

Tx sinsinmax

(a) sm3.1

Hz100.12m1020.02 33max

fAAv

(b)The maximum speedoccurs midway betweenthe ends of its motion.

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Simple harmonic motion?

• Is the motion of the lighted bulb simple harmonic motion, when each lights for 0.5s in sequence?

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10.2 Simple Harmonic Motion and the Reference Circle

ACCELERATION

tAaaa

cx coscosmax

2

The ball on the reference circle moves in uniform circular motion, and, therefore, has centripetal acceleration ac that points toward the centre of the circle.

The acceleration ax of the shadow is the x component of the centripetal acceleration ac .

2rac

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Maximum Acceleration

• A loudspeaker vibrating at 1kHz with an amplitude of 0.20mm has a maximum acceleration of

amax = A2 = 7.9 x 103 m/s2

• Maximum acceleration occurs when the force acting on the diaphragm is a maximum.

• The maximum force arises when the diaphragm is at the ends of its path.

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Frequency of Vibration

• With the aid of the Newton’s second law , it is possible to determine the frequency at which an object of mass m vibrates on a spring.

• Mass of the spring is negligible

• Only force acting is the restoring force.

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10.2 Simple Harmonic Motion and the Reference Circle

FREQUENCY OF VIBRATION

m

k

tAax cos2tAx cos

xmakxF 2mAkA

Larger spring constants kand smaller masses mresult in larger frequencies.

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10.2 Simple Harmonic Motion and the Reference Circle

Example 6 A Body Mass Measurement Device

The device consists of a spring-mounted chair in which the astronautsits. The spring has a spring constant of 606 N/m and the mass ofthe chair is 12.0 kg. The measured period is 2.41 s. Find the mass of theastronaut.

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10.2 Simple Harmonic Motion and the Reference Circle

totalm

k 2

total km

Tf

22

astrochair2total2

mmT

km

kg 77.2kg 0.124

s 41.2mN606

2

2

2

chair2astro

m

T

km

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10.3 Energy and Simple Harmonic Motion

A compressed spring can do work.

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Average magnitude of Force

Spring force at x0 is kx0

Spring force at xf is kxf

Average Fx = ½(kx0+kxf)

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10.3 Energy and Simple Harmonic Motion

fofo xxkxkxsFW 0coscos 21

elastic

2212

21

elastic fo kxkxW

Work done by the average spring force:

Initial elastic PE Final elastic PE

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10.3 Energy and Simple Harmonic Motion

DEFINITION OF ELASTIC POTENTIAL ENERGY

The elastic potential energy is the energy that a springhas by virtue of being stretched or compressed. For anideal spring, the elastic potential energy is

221

elasticPE kx

SI Unit of Elastic Potential Energy: joule (J)

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10.3 Energy and Simple Harmonic Motion

Conceptual Example 8 Changing the Mass of a Simple Harmonic Oscilator

The box rests on a horizontal, frictionlesssurface. The spring is stretched to x=Aand released. When the box is passingthrough x=0, a second box of the samemass is attached to it. Discuss what happens to the (a) maximum speed(b) amplitude (c) angular frequency.

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Doubling the Mass of a Simple Harmonic Oscillator

• At x=0m, a second box of the same mass and speed vmax is attached.

• So, the max KE is doubled as mass is doubled.

• So, when the spring compresses it will have double the PEe .

• As PEe is doubled max amplitude will be 2 times

221

elasticPE kx

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Doubling the Mass causes max amplitude to 2 times

• Before adding the second mass, the displacement x1 is related to the PE as:

• As PEe is doubled the new amplitude is x2

212

1elasticPE kx

222

1elastic2PE kx

12

21

22

21

222

1212

1

2

2

2

2

xx

xx

xx

kxkx

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Doubling the Mass of a Simple Harmonic Oscillator

Angular frequency is

So, a doubling of mass will cause the angular frequency reduce to

(1/ 2 )

m

k

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10.3 Energy and Simple Harmonic Motion

Example 9 A falling ball on a vertical Spring

A 0.20-kg ball is attached to a vertical spring. The spring constantis 28 N/m. When released from rest, how far does the ball fallbefore being brought to a momentary stop by the spring?

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10.3 Energy and Simple Harmonic Motion

of EE

2212

212

212

212

212

21

ooooffff kymghImvkymghImv

oo mghkh 221

m 14.0

mN28

sm8.9kg 20.02

2

2

k

mgho

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10.4 The Pendulum

A simple pendulum consists of a particle mass m attached to a frictionlesspivot by a cable of negligible mass.

only) angles (small L

g

only) angles (small I

mgL

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simple pendulumThe force of gravity is responsible for the

back-and-forth rotation about the axis at P.

A net torque is required to change the angular speed. The gravitational force mg produces the torque.

= - (mg)l - sign to represent the restoring force

= - (mg)L = - k’ where k’ = mgL Same as F = - kx

So, = (k’/m) = (mgL/m) = (mgL/I) in rotational motion in place of mass moment of inertia ‘I’ will appear

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simple pendulumSo, = (k’/m) = (mgL/m) = (mgL/I)

in rotational motion in place of mass moment of inertia ‘I’ will appear

The moment of inertia of a particle of mass m, rotating at a radius L about an axis is

I = mL2

Frequency of a simple pendulum is

Which does not depend on the mass of the particle.

only) angles (small L

g

PHYS222 – LSSU – Bazlur Slide 43

10.4 The Pendulum

Example 10 Keeping Time

Determine the length of a simple pendulum that willswing back and forth in simple harmonic motion with a period of 1.00 s.

2

2L

g

Tf

m 248.0

4

sm80.9s 00.1

4 2

22

2

2

gTL

2

2

4gT

L

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10.5 Damped Harmonic Motion

In simple harmonic motion, an object oscillated with a constant amplitude.

In reality, friction or some other energy dissipating mechanism is always present and the amplitude decreases as time passes.

This is referred to as damped harmonic motion.

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10.5 Damped Harmonic Motion

1) simple harmonic motion

2 & 3) underdamped

4) critically damped

5) overdamped

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10.6 Driven Harmonic Motion and Resonance

When a force is applied to an oscillating system at all times,the result is driven harmonic motion.

Here, the driving force has the same frequency as the spring system and always points in the direction of the object’s velocity.

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10.6 Driven Harmonic Motion and Resonance

RESONANCE

Resonance is the condition in which a time-dependent force can transmitlarge amounts of energy to an oscillating object, leading to a large amplitudemotion.

Resonance occurs when the frequency of the force matches a natural frequency at which the object will oscillate.

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Elastic Deformation

• All materials become distorted when they are squeezed or stretched.

• Those materials return to their original shape when the deforming force is removed, such materials are called “elastic”.

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10.7 Elastic Deformation

Because of these atomic-level “springs”, a material tends to return to its initial shape once forces have been removed.

ATOMS

FORCES

Atomic View of Elastic materials

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Stretching force

• Magnitude of the deforming force can be expressed as follows, provided the amount of stretch or compression is small compared to the original length of the object.

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10.7 Elastic Deformation

STRETCHING, COMPRESSION, AND YOUNG’S MODULUS

AL

LYF

o

Y is the proportionality constant called Young’s modulus

Young’s modulus has the units of pressure: N/m2

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Young’s modulus• The magnitude of the force is proportional to the

fractional increase (or decrease) in length L/L0, rather than the absolute change L.

• Young’s modulus depends on the nature of the material.

• For a given force, the material with higher Y undergoes the smaller change in length.

AL

LYF

o

This force also called the “tensile” force,

because they cause a tension in the material

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10.7 Elastic Deformation

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10.7 Elastic Deformation

Example 12 Bone Compression

In a circus act, a performer supports the combined weight (1080 N) ofa number of colleagues. Each thighbone of this performer has a length of 0.55 m and an effective cross sectional area of 7.7×10-4 m2. Determinethe amount that each thighbone compresses under the extra weight.

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10.7 Elastic Deformation

AL

LYF

o

m 101.4

m 107.7mN104.9

m 55.0N 540 52429

YA

FLL o

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Shear Deformation

• It is possible to deform a solid object in a way other than stretching or compressing it.

• When a top cover of a book is pushed the pages below it become shifted relative to the stationary bottom cover.

• The resulting deformation is called a shear deformation.

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Shear Deformation• When a force F is applied parallel to the surface of an object of

area A, the object experiences a shear x (the top surface moves relative to the bottom surface) for an object with thickness L0.

• The magnitude of the shearing force is proportional to the fractional shear x in thickness x/L0, rather than the absolute change x.

AL

xSF

o

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Shear Modulus

• The constant of proportionality S is called the shear modulus.

• Shear modulus depends on the nature of the material• For a given force, the material with higher S

undergoes the smaller shear x .• Shear modulus has the units of pressure: N/m2

AL

xSF

o

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Shear Deformation And The Shear Modulus

AL

LYF

o

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10.7 Elastic Deformation

SHEAR DEFORMATION AND THE SHEAR MODULUS

AL

xSF

o

The shear modulus has the units of pressure: N/m2

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10.7 Elastic Deformation

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10.7 Elastic Deformation

Example 14: Shear Modulus of J-E-L-L-O

You push tangentially across the topsurface with a force of 0.45 N. The top surface moves a distance of 6.0 mmrelative to the bottom surface. What isthe shear modulus of Jell-O?

AL

xSF

o

xA

FLS o

PHYS222 – LSSU – Bazlur Slide 63

10.7 Elastic Deformation

2

32 mN460m 100.6m 070.0

m 030.0N 45.0

S

xA

FLS o

Shear Modulus of J-E-L-L-O

Jell-o can be deformed easily, because its Shear Modulus is significantly ___________ than that of other rigid materials, e.g., steel

PHYS222 – LSSU – Bazlur Slide 64

Young’s modulus vs. Shear modulus

AL

xSF

o

A

L

LYF

o

• Young’s modulus (Y) refers to the change in length of one dimension of a solid object as a result of tensile or compressive force.

• The shear modulus (S) refers to a change in shape of a solid object as a result of shearing force.

PHYS222 – LSSU – Bazlur Slide 65

Young’s modulus vs. Shear modulus

AL

xSF

o

A

L

LYF

o

• The tensile force is perpendicular to the surface

of area A• The shear force is parallel to the surface

of area A

• The distances L and L0 (length) are parallel

• The distances x and L0 (thickness) are perpendicular

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Volume Deformation• When applied compressive forces

changes the size of every dimension (length, width, and depth), leading to a decrease in volume, the resulting deformation is called a volume deformation.

• This kind of overall compression occurs when an object is submerged in a liquid.

• The force acting in such situation s are applied perpendicular to every surface.

• The perpendicular force per unit area is pressure.

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Pressure• The pressure P is the magnitude of the force F acting

perpendicular to a surface divided by the area A over which the surface acts:

• SI unit of Pressure is Pa (pascal) = N/m2

A

FP F

A

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Volume DeformationThe change in pressure P needed

to change the volume by an amount V is directly proportional to the fractional change V/V0 in volume:

Change in pressure P = final pressure P - initial pressure P0

Change in volume v = final volume V - initial volume V0

oV

VBP

PHYS222 – LSSU – Bazlur Slide 69

10.7 Elastic Deformation

VOLUME DEFORMATION AND THE BULK MODULUS

oV

VBP

The Bulk modulus has the units of pressure: N/m2

B is the proportionality constant called Bulk modulus

The negative sign represents the increase in pressure decreases the volume

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10.7 Elastic Deformation

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Stress, Strain, and Hooke’s Law• All these modified equations

specify the amount of force needed per unit area for a given amount of elastic deformation.

• In general, the ratio of the magnitude of the force to the area is called the stress.

• The right side of each equation involves the change in a quantity (L, X, or V) divided by a quantity (L or V) relative to which the change is compared. Each of these ratios are referred to as the strain that results from the stress.

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Stress, Strain, and Hooke’s Law• Stress and strain are

directly proportional to one another.

• This relationship was first discovered by Robert Hooke and referred to a Hooke’s Law.

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10.8 Stress, Strain, and Hooke’s Law

HOOKE’S LAW FOR STRESS AND STRAIN

Stress is directly proportional to strain.

Strain is a unitless quantitiy.

SI Unit of Stress: N/m2

In general the quantity F/A is called the stress.

The change in the quantity divided by that quantity is called thestrain:

ooo LxLLVV

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10.8 Stress, Strain, and Hooke’s Law

•In reality, materials obey Hooke’s law only up to a certain limit, proportionality limit. •Elastic limit is the point beyond which the object no longer returns to its original size and shape when the stress is removed.

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ForceA

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ForceA

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ForceA

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ForceA