PHYSICS 231

INTRODUCTORY PHYSICS I

Lecture 20

Heat Engine Refrigerator,

Heat Pump

Last Lecture

Qhot

engine

Qcold

W

Qhot

fridge

Qcold

W

€

e = WQhot

< eCarnot =1−TcoldThot

€

W =Qhot −Qcold€

COP =Qc,hW

< COPCarnot =Tc,h

Thot −Tcold

Entropy• Measure of Disorder of the system

(randomness, ignorance)• Entropy: S = kBlog(N)

N = # of possible arrangements for fixed E and Q

0

100

200

300

400

500

600

700

800

900

1000

(0,12) (1,11) (2,10) (3,9) (4,8) (5,7) (6,6) (7,5) (8,4) (9,3) (10,2) (11,1) (12,0)

Number of ways for 12 molecules to arrange themselves in two halves of container.

S is greater if molecules spread evenly in both halves.

On a macroscopic level, one finds that adding heat raises entropy:

Temperature in Kelvin!

ΔS =Q /T

2nd Law of Thermodynamics(version 2)

The Total Entropy of the Universe can never decrease.

(but entropy of system can increase or decrease)

Why does Q flow from hot to cold?

• Consider two systems, one with TA and one with TB

• Allow Q > 0 to flow from TA to TB

• Entropy changes by:

ΔS = Q/TB - Q/TA

• This can only occur if ΔS > 0, requiring TA > TB.

• System will achieve more randomness by exchanging heat until TB = TA

Carnot Engine

Carnot cycle is most efficient possible, because the total entropy change is zero. It is a “reversible process”.

For real engines:

€

ΔS = ΔSenvironment =QcoldTcold

− QhotThot

> 0 ⇒

€

e = WQhot

=1− QcoldQhot

<1− TcoldThot

= eCarnot

Chapter 13

Vibrations and Waves

• When x is positive ,F is negative ;

• When at equilibrium (x=0), F = 0 ;

• When x is negative ,F is positive ;

Hooke’s Law Reviewed

F =−kx

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are needed to see this picture.

Sinusoidal Oscillation

If we extend the mass, and let go, the pen traces a sine wave.

Graphing x vs. t

A : amplitude (length, m) T : period (time, s)

A

T

Amplitude: A

Period: T

Frequency: f = 1/T

Angular frequency:

A

T

Period and Frequency

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x = Acosωt

€

T = 2π⇓

€

T = 2πω, f = ω

2π

Phases

Often a phase is included to shift the timing of the peak:

Phase of 90-degrees changes cosine to sine

cos t−π2

⎛⎝⎜

⎞⎠⎟ =sin t( )

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x = Acos ωt −φ( )= Acos ω(t − t0)( ) for peak at

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t = t0

a

x

v• Velocity is 90 “out of phase” with x: When x is at max,v is at min ....

• Acceleration is 180° “out of phase” with x a = F/m = - (k/m) x

Velocity and Acceleration vs.

time

T

T

T

v and a vs. t

Find vmax with E conservation

Find amax using F=ma

x =Acostv=−vm ax sinta=−am ax cost

12kA2 =

12mvm ax

2

vm ax =Akm

−kx = ma−kAcosωt = −mamax cosωt

amax = Akm

Connection to Circular Motion

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are needed to see this picture.

circular motion with constant angular velocity

Simple Harmonic Motion

Projection on axis

What is ?

=kmQuickTime™ and a

Animation decompressorare needed to see this picture.

Circular motion

Angular speed: Radius: A

=> Speed: v=A

Simple Harmonic Motion

Cons. of E:

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vmax = Akm

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=km

Formula Summary

f =1T

=2π =2πT

x =Acos(t−φ)v=−Asin(t−φ)

a=− 2A(cost−φ)

=km

Example13.1

An block-spring system oscillates with an amplitude of 3.5 cm. If the spring constant is 250 N/m and the block has a mass of 0.50 kg, determine

(a) the mechanical energy of the system

(b) the maximum speed of the block

(c) the maximum acceleration.

a) 0.153 J

b) 0.783 m/s

c) 17.5 m/s2

Example 13.2

A 36-kg block is attached to a spring of constant k=600 N/m. The block is pulled 3.5 cm away from its equilibrium positions and released from rest at t=0. At t=0.75 seconds,

a) what is the position of the block?

b) what is the velocity of the block?

a) -3.489 cm

b) -1.138 cm/s

Example 13.3

A 36-kg block is attached to a spring of constant k=600 N/m. The block is pulled 3.5 cm away from its equilibrium position and is pushed so that is has an initial velocity of 5.0 cm/s at t=0.

a) What is the position of the block at t=0.75 seconds?

a) -3.39 cm

Example 13.4aAn object undergoing simple harmonic motion follows the expression,

x(t)=4 + 2cos[π(t−3)]

The amplitude of the motion is:a) 1 cmb) 2 cmc) 3 cmd) 4 cme) -4 cm

Where x will be in cm if t is in seconds

Example 13.4bAn object undergoing simple harmonic motion follows the expression,

x(t)=4 + 2cos[π(t−3)]

The period of the motion is:a) 1/3 sb) 1/2 sc) 1 sd) 2 se) 2/π s

Here, x will be in cm if t is in seconds

Example 13.4cAn object undergoing simple harmonic motion follows the expression,

x(t)=4 + 2cos[π(t−3)]

The frequency of the motion is:a) 1/3 Hzb) 1/2 Hzc) 1 Hzd) 2 Hze) π Hz

Here, x will be in cm if t is in seconds

Example 13.4dAn object undergoing simple harmonic motion follows the expression,

x(t)=4 + 2cos[π(t−3)]

The angular frequency of the motion is:a) 1/3 rad/sb) 1/2 rad/sc) 1 rad/sd) 2 rad/se) π rad/s

Here, x will be in cm if t is in seconds

Example 13.4eAn object undergoing simple harmonic motion follows the expression,

x(t)=4 + 2cos[π(t−3)]

The object will pass through the equilibrium positionat the times, t = _____ seconds

a) …, -2, -1, 0, 1, 2 … b) …, -1.5, -0.5, 0.5, 1.5, 2.5, …c) …, -1.5, -1, -0.5, 0, 0.5, 1.0, 1.5, …d) …, -4, -2, 0, 2, 4, …e) …, -2.5, -0.5, 1.5, 3.5,

Here, x will be in cm if t is in seconds