To understand elasticity and plasticity. To understand ...people.physics.tamu.edu/wwu/docs/P201S2017NotesChapter11.pdf · To understand elasticity and plasticity. To understand simple

● To understand stress, strain, and elastic deformation.● To understand elasticity and plasticity.● To understand simple harmonic motion (SHM).● To solve equations of simple harmonic motion.● To understand the pendulum as a an example of SHM.

Chapter 11 Elasticity and Periodic Motion

11.1 Stress, Strain, and Elastic Deformation

What is stress?

The “intensity” of the force exerting on an object

quantified by the force per unit area.

What is strain?

The amount of relative deformation appears to an object

under the given stress.

The relationship between the two----If the stress is small,

the resultant strain is proportional to the stress:

𝑆𝑡𝑟𝑒𝑠𝑠

𝑆𝑡𝑟𝑎𝑖𝑛= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

Tensile and Compressive Stress and Strain

Tensile and compressive stress

Tensile stress = 𝐹⊥

𝐴

Units: N/m2 or pascal (Pa), in SI unit system

psi or pounds per square inches, in the British units

Tensile and compressive strain

Tensile strain = 𝑙−𝑙0

𝑙0=

∆𝑙

𝑙0

Units: none

Young’s modulus

𝑌 =𝑇𝑒𝑛𝑠𝑖𝑙𝑒 𝑆𝑡𝑟𝑒𝑠𝑠

𝑇𝑒𝑛𝑠𝑖𝑙𝑒 𝑆𝑡𝑟𝑎𝑖𝑛=

𝐹⊥/𝐴

∆𝑙/𝑙0=

𝑙0

𝐴

𝐹⊥

∆𝑙

Units: N/m2 or Pa

Example 11.1 on page 324

A stretching elevator cable

Given: m, 𝑙0, A, and ∆𝑙

Find: the cables stress, strain, and Young’s modulus

Solution:

Stress =𝐹⊥

𝐴=

𝑊

𝐴=

𝑚𝑔

𝐴=

(550 𝑘𝑔)(9.8 𝑚/𝑠2)

0.20×10−4𝑚2 = 2.7 × 108 Pa

Strain = 𝑙−𝑙0

𝑙0=

∆𝑙

𝑙0=

0.40×10−2𝑚

3.0 𝑚= 0.00133

Young’s Modulus

𝑌 =𝑆𝑡𝑟𝑒𝑠𝑠

𝑆𝑡𝑟𝑎𝑖𝑛=2.7 × 108 Pa

0.00133= 2.0 × 1011 𝑃𝑎

Volume Stress and Strain

Volume stress, or, pressure

𝑝 =𝐹⊥

𝐴

Units: N/m2 or Pa, in SI units

psi, in the British units

Volume strain

Volume strain = 𝑉−𝑉0

𝑉0=

∆𝑉

𝑉0

Units: none

Bulk modulus

𝐵 =𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒

𝑟𝑒𝑠𝑢𝑙𝑡𝑖𝑛𝑔 𝑣𝑜𝑙𝑢𝑚𝑒 𝑠𝑡𝑟𝑎𝑖𝑛=

∆𝑝

∆𝑉/𝑉0

Units: N/m2 or Pa

Shear Stress and Strain

Shear stress

Shear stress = 𝐹∥

𝐴

Units: N/m2 or Pa

Shear strain

Shear strain = 𝑥

ℎ= tan𝜙 ≈ 𝜙 for x ≪ h

Units: none

Shear modulus

𝑆 =𝑆ℎ𝑒𝑎𝑟 𝑆𝑡𝑟𝑒𝑠𝑠

𝑆ℎ𝑒𝑎𝑟 𝑆𝑡𝑟𝑎𝑖𝑛=

𝐹∥/𝐴

𝑥/ℎ=

𝐹∥/𝐴

𝜙

Units: N/m2 or Pa

Elasticity and Plasticity

11.2 Periodic Motion

Simple Harmonic Motion (SHM) illustrated by the oscillations

of a mass-loaded spring

Spring restoring force: 𝐹𝑥 = −𝑘𝑥

Acceleration of the mass: 𝑎𝑥 =𝐹𝑥

𝑚= −

𝑘

𝑚𝑥

Defining simple harmonic motion: motion driven by a

restoring force that is always opposite to the displacement and

directly proportional to the displacement in magnitude.

Note:

(a) The restoring force Fx is opposite to displacement;

(b) Fx is not a constant;

(b) As a result, the acceleration ax is not a constant;

(c) ax varies between (+ kA/m) and (– kA/m);

(d) The magnitude of ax has a maximum amax = kA/m.

Quantities that Describe Periodic Motion

● The amplitude of the motion is the maximum magnitude of the displacement,

𝐴 = |𝑥|𝑚𝑎𝑥.

● One cycle of the motion: one complete round trip.

● The period T of the motion is the time it takes to complete one cycle, in units of s.

● The frequency f of the motion is the number of cycles the motion complete in 1 s.

● The angular frequency 𝜔 of the motion is 2𝜋 multiplies the frequency, 𝜔 = 2𝜋𝑓.

● The relationships between T, f, and 𝜔:

𝑓 =1

𝑇Units: hertz or Hz 1 Hz = 1 s-1

𝜔 = 2𝜋𝑓 =2𝜋

𝑇Units: rad/s

11.3 Energy in Simple Harmonic Motion

Conservation of Energy in SHM

𝐸 =1

2𝑘𝐴2 =

1

2𝑚𝑣𝑥

2 +1

2𝑘𝑥2

Velocity of an object in SHM as a function of position

𝑣𝑥 = ±𝑘

𝑚(𝐴2 − 𝑥2) = ±

𝑘

𝑚(𝐴2 − 𝑥2)

Note:

(a) 𝑣𝑥 = 0 when 𝑥 = ±𝐴.

(b) Maximum speed 𝑣𝑥,𝑚𝑎𝑥 = 𝐴 𝑘/𝑚 when 𝑥 = 0.

Example 11.5 on page 334, SHM on an air track

Given: m and A. Force constant k is give indirectly.

Find: (a) force constant k

(b) the maximum & minimum velocities

(c) the maximum & minimum accelerations

(d) v & a at half way to x = 0

(e) K, U, and E at half way to x = 0

(a) Force constant k = F/x = (6.0 N)/(0.030 m) = 200 N/m

(b) 𝑣𝑚𝑎𝑥 = 𝐴 𝑘/𝑚 = 0.80 m/s, 𝑣𝑚𝑖𝑛 = −𝐴 𝑘/𝑚 = − 0.80 m/s, both at 𝑥 = 0

(c) amax = + kA/m = 16.0 m/s2 at 𝑥 = − 0.04 m; amin = − kA/m = − 16.0 m/s2 at 𝑥 = + 0.04 m

(d) 𝑣𝑥 = −𝑘

𝑚𝐴2 − 𝑥2 = −

𝑘

𝑚𝐴2 −

1

2𝐴

2= − 0.69 m/s

𝑎𝑥 = −𝑘

𝑚𝑥 = −

𝑘

𝑚

1

2𝐴 = − 8.0 m/s2

(e) 𝐾 =1

2𝑚𝑣𝑥

2 = 0.12 J 𝑈 =1

2𝑘𝑥2 = 0.040 J 𝐸 = 𝐾 + 𝑈 = 0.16 J

11.4 Equation of Simple Harmonic Motion

Circle of Reference: Consider a small object undergoing a uniform

circular motion as shown in the sketch. The x-component of the position

of the object obeys SHM exactly. Therefore, the expressions describing

the x-component of its positon can be used to describe SHM.

Let the angular velocity of the object be 𝜔, and, its angular position

𝜙𝑜 = 0 at 𝑡 = 0. Let the radius of the circle be A.

The object’s the angular position at time 𝑡 is 𝜙 = 𝜔𝑡. We have the

following x-component quantities for the object:

The position 𝑥 = 𝐴𝑐𝑜𝑠 𝜙 = 𝐴𝑐𝑜𝑠(𝜔𝑡) …..Equation of SHM

The velocity 𝑣𝑥 = − 𝑣 𝑠𝑖𝑛 𝜙 = −(𝜔𝐴)𝑠𝑖𝑛(𝜔𝑡)

The acceleration 𝑎𝑥 = − 𝑎𝑟𝑎𝑑 𝑐𝑜𝑠 𝜙 = −𝜔2𝐴 𝑐𝑜𝑠 𝜔𝑡 = −𝜔2𝑥

What is 𝜔? Since for the spring along the x-axis 𝑎𝑥 = Fx/m = − k𝑥/m

we have 𝜔2 = 𝑘/𝑚 or 𝜔 = 𝑘/𝑚,

and other quantities 𝑓 =𝜔

2𝜋=

1

2𝜋𝑘/𝑚, 𝑇 =

1

𝑓= 2𝜋 𝑚/𝑘

Graphic Description of Position, Velocity, and Acceleration

The position: 𝑥 = 𝐴𝑐𝑜𝑠 𝜙 = 𝐴𝑐𝑜𝑠(𝜔𝑡)

The velocity: 𝑣𝑥 = − 𝑣 𝑠𝑖𝑛 𝜙 = −(𝜔𝐴)𝑠𝑖𝑛(𝜔𝑡)

The acceleration: 𝑎𝑥 = − 𝑎𝑟𝑎𝑑 𝑐𝑜𝑠 𝜙 = −𝜔2𝐴 𝑐𝑜𝑠 𝜔𝑡

= −𝜔2𝑥

Parameters:

𝜔2 = 𝑘/𝑚 or 𝜔 = 𝑘/𝑚,

𝑓 =𝜔

2𝜋=

1

2𝜋𝑘/𝑚

𝑇 =1

𝑓= 2𝜋 𝑚/𝑘

A few notes about SHM:

● The angular frequency, period, and frequency are all independent of the amplitude.

𝜔 = 𝑘/𝑚; 𝑇 =1

𝑓= 2𝜋 𝑚/𝑘; 𝑓 =

𝜔

2𝜋=

1

2𝜋𝑘/𝑚

● If 𝜙𝑜 ≠ 0 at 𝑡 = 0, the previously derived expressions remain correct if the angular

position 𝜙 = 𝜔𝑡 is replaced by 𝜙 = 𝜙𝑜 +𝜔𝑡. For example, the position is

𝑥 = 𝐴𝑐𝑜𝑠 𝜙 = 𝐴𝑐𝑜𝑠(𝜙𝑜 + 𝜔𝑡)

● Since 𝑥 = 𝐴𝑐𝑜𝑠 𝜙 , we may re-write

𝑣𝑥 = −𝜔𝐴𝑠𝑖𝑛 𝜔𝑡 = ±𝜔𝐴 1 − cos 𝜔𝑡 2

= ±𝜔 𝐴2 − 𝐴 cos 𝜔𝑡 2 = ±𝜔 𝐴2 − 𝑥2

= ±𝑘

𝑚(𝐴2 − 𝑥2)

11.5 The Simple Pendulum

The mass moves along a circular path but the tangential speed

is not a constant. Define 𝜃 = 0 as the vertical and counter-

clockwise as positive. Define x as the position of the mass

along the arc in a similar way.

The tangential force is 𝐹𝑥 = −𝑚𝑔𝑠𝑖𝑛(𝜃).

In the small angle limit 𝐹𝑥 ≈ −𝑚𝑔𝜃 = −𝑚𝑔𝑥

𝐿.

Newton’s Second Law −𝑚𝑔𝑥

𝐿= 𝑚𝑎𝑥 or −

𝑔

𝐿𝑥 = 𝑎𝑥

Compared with the SHM of a mass-loaded spring,

Spring restoring force 𝐹𝑥 = −𝑘𝑥.

Newton’s Second Law − 𝑘𝑥 = 𝑚𝑎𝑥 or −𝑘

𝑚𝑥 = 𝑎𝑥

Conclusions for the simple pendulum:

𝜔 = 𝑔/𝐿; 𝑇 =1

𝑓= 2𝜋 𝐿/𝑔; 𝑓 =

𝜔

2𝜋=

1

2𝜋𝑔/𝐿