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Period T the number of cycles per second (Hz) T f 1 T f 2 2 Amplitude A the maximum displacement (m) the time required to complete one cycle (s) Frequency f Angular frequency rad/s t A A x cos cos Summary of Last Class HOOKE’S LAW: x k F x t A v v sin max t A a a cos max 2 = − 2 OR m k For a mass m attached to a spring and set in vibration on frictionless horizontal surface 2 2 1 elastic PE kx joule (J)

Summary of Last Class HOOKE’S LAW v A t Z sin Z

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Period T

the number of cycles per second (Hz)

Tf

1

Tf

22

Amplitude A the maximum displacement (m)

the time required to complete one cycle (s)

Frequency f

Angular frequency rad/s

tAAx coscos

Summary of Last Class

HOOKE’S LAW: xkFx

tAvv

sinmax

tAaa

cosmax

2 𝑎 = −𝜔2𝑥 OR

m

k

For a mass m attached to

a spring and set in

vibration on frictionless

horizontal surface

2

21

elasticPE kx joule (J)

10.2 Simple Harmonic Motion and the Reference Circle

Example: 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?

tAvvv

Tx sinsinmax

f 2

𝑣𝑚𝑎𝑥 =?

Where?

6283.2 rad/s

1.26 𝑚/𝑠

When passing through

equilibrium, x = 0

𝑣𝑚𝑎𝑥 = 𝐴𝜔

10.3 Energy and Simple Harmonic Motion

Example: Adding a Mass to a Simple Harmonic Oscillator

A 0.20-kg ball is attached to a vertical spring. The spring constant

is 28 N/m. When released from rest, how far does the ball fall

before being brought to a momentary stop by the spring?

𝑥 = 0.07 𝑚

𝑘𝑥 = 𝑚𝑔

10.3 Energy conservation

of EE

2

212

212

212

212

212

21

ooooffff kxmghImvkxmghImv

𝐸 = 𝐾𝐸𝑇 + 𝐾𝐸𝑅 + 𝑃𝐸𝑔 + 𝑃𝐸𝑒

In absence of any external force

Principle of Energy

Conservation

Simple Pendulum

𝜃 𝐿

𝑥

𝑚𝑔

𝜏 = 𝐼𝛼 𝐼 = 𝑚𝑅2

𝐼 = 𝑚𝐿2 𝜏 = −𝑚𝑔𝑥

𝑎𝑇 = 𝐿𝛼 𝛼 =𝑎𝑇𝐿

−𝑚𝑔𝑥 = 𝑚𝐿2𝑎𝑇𝐿

𝑎𝑇 = −𝐴𝜔2 cos 𝜔𝑡 = −𝜔2𝑥

𝜔 =𝑔

𝐿

𝑥

𝜔 = 2𝜋𝑓 𝑓 =𝜔

2𝜋=1

2𝜋

𝑔

𝐿

𝑇 =1

𝑓 𝑇 = 2𝜋

𝐿

𝑔 Period of a simple pendulum is

independent of mass and

Amplitude of the pendulum

As 𝑎𝑇 = 𝑟𝛼

Q: A bob attached to a string and set into simple harmonic (a simple

pendulum is when angle of swing remains small). One such simple pendulum

has a period equals to 0.1 s. Now if the bob is changed to a slightly bigger one

with mass double than the previous bob, keeping length of the string same,

the period of the simple pendulum will

Q: A simple pendulum has a period of 6.0 s on the surface of earth. The

period of the same pendulum on the surface of moon (where the

acceleration due to gravity is 1/6 of that on the surface of earth) will

(c) Remain unchanged (a) Become double (b) Become half

(a) Be the same (b) Increase (c) Decrease

𝑇 = 2𝜋𝐿

𝑔

Problem: A simple pendulum has a ball of mass m attached to a

string of length 1.50 m. The ball is pulled to one side through a small

angle and then released from rest.

(a) After the ball is released how much time is elapsed before it

gains its maximum speed?

(b) After the ball is released how much time is elapsed before it gains

its maximum acceleration?

(c) What will be speed of the ball at the time when the acceleration

is maximum?

(d) What is the angular frequency of this simple pendulum?

(e) What is the linear frequency of this simple pendulum?

Simple pendulum:

0.62 s

1.23 s

0

0.41 Hz

f (Hz) = 1/T 𝜔 =𝑔

𝐿 T = 2π √(L/g)

2.56 rad/s

10.7 Elastic Deformation

STRETCHING, COMPRESSION, AND YOUNG’S MODULUS

AL

LYF

o

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

Table 10.1: Young Modulus of Elasticity for some materials

Q: With same applied force rubber can be elastically deformed

thousands of time more than steel. Does it mean that the Young’s

modulus of elasticity of rubber is

(a) Thousands of time larger than that of steel?

(b) Thousands of time smaller than that of the steel?

(c) Same as that of steel.

10.7 Elastic Deformation

SHEAR DEFORMATION AND THE SHEAR MODULUS

AL

xSF

o

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

Deformation is perpendicular to original length

Table 10.2: Shear Modulus of Elasticity for some materials

10.7 Elastic Deformation

Example 14 J-E-L-L-O

You push tangentially across the top

surface with a force of 0.45 N. The

top surface moves a distance of 6.0 mm

relative to the bottom surface. What is

the shear modulus of Jell-O?

AL

xSF

o

xA

FLS o

𝐿0 = .03 𝑚 𝐴 = .07 × .07 𝑚2 ∆𝑥 = .006 𝑚

459 N/m2

10.7 Elastic Deformation

VOLUME DEFORMATION AND THE BULK MODULUS

oV

VBP

The Bulk modulus has the

units of pressure: N/m2

Table 10.3: Bulk Modulus of Elasticity for some materials

10.7 Elastic Deformation

10.8 Stress, Strain, and Hooke’s Law

HOOKE’S LAW FOR STRESS AND STRAIN

In general the quantity 𝐹

𝐴 is called the Stress

oLx

N/m2

oLL

SI Unit of Stress:? SI Units of

Strain?

Stress is directly proportional to strain.

oVV

oL

LY

A

F

oL

xS

A

F

oV

VBP

A

FP

The change in the dimension divided by that original is called the Strain

STRETCHING, COMPRESSION SHEAR BULK

Elastic Deformations

A unit less quantity

For Recitation Practice

Chapter 10

FOC: 2, 4, 11 & 14.

Problems: 5, 11, 17, 27 & 51.

Reading Assignment for Next class

Ch 10: 10.4, 10.7 and 10.8