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Physics 2053C – Fall 2001. Chapter 11 Simple Harmonic Motion. T H. Q H. W. Q L. Engine. T L. Carnot Cycle  The most efficient Engine Possible. Q = U + W  = W/Q in = (Q H -Q L )/(Q H ) = (T H -T L )/(T H )  = 1 – (T L /T H ). T H. Q H. W. Q L. Engine. T L.

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physics 2053c fall 2001

Physics 2053C – Fall 2001

Chapter 11

Simple Harmonic Motion

Dr. Larry Dennis,

FSU Department of Physics

carnot cycle the most efficient engine possible

TH

QH

W

QL

Engine

TL

Carnot Cycle  The most efficient Engine Possible
  • Q = U + W
  •  = W/Qin

= (QH-QL)/(QH)

= (TH-TL)/(TH)

 = 1 – (TL/TH )

carnot cycle the most efficient engine possible1

TH

QH

W

QL

Engine

TL

Carnot Cycle  The most efficient Engine Possible

Example: An engine absorbs 200 J of heat and does 75 J of work per cycle.

What is it’s efficiency?

How much heat does it eject into the cold reservoir?

If the temperature of the cold reservoir is 190 K, what is the temperature of the hot reservoir?

If the engine undergoes 2500 cycles per second, what is it’s power?

carnot cycle the most efficient engine possible2

TH

QH

W

QL

Engine

TL

Carnot Cycle  The most efficient Engine Possible

What is it’s efficiency?

 = Work/Heat In

= 75 J/200 J = .375

How much heat does it eject into the cold reservoir?

Q = U + W

QH – QL = U + W

QL = QH – W = 200 – 75 J = 125 J

carnot cycle the most efficient engine possible3

TH

QH

W

QL

Engine

TL

Carnot Cycle  The most efficient Engine Possible

If the temperature of the cold reservoir is 190 K, what is the temperature of the hot reservoir?

= 1 – TL/TH = 0.375

TH – TL = 0.375 * TH

TL = TH – 0.375 * TH

TL = 0.625*TH

TH = TL/0.625

= 190 / 0.625 = 304 K

carnot cycle the most efficient engine possible4

TH

QH

W

QL

Engine

TL

Carnot Cycle  The most efficient Engine Possible

If the engine undergoes 2500 cycles per second, what is it’s power?

P = W/t = 2500 * 75 J/1 s

P = 187500 Watts

Note this is about 250 h.p.

periodic motion
Periodic Motion
  • Many kinds of motion are periodic.
  • Examine within the context of:
    • Forces
    • Energy
  • Equilibrium Position & Restoring Forces
position vs time
Position vs. Time

x = A sin(2*t/T)

Position = Amplitude * sin( 2 * time/Period)

x

Amplitude = A

t

Period = T

spring oscillations

 = 2f = 2/T = (k/m)

Spring Oscillations

x

k

m

x(t) = Acos(t) or Asin(t)

pendulum oscillations

T

Net Force

mg

 = 2f = 2/T = (g/L)

Pendulum Oscillations

L

(t) = A cos(t) or (t) = A sin(t)

Note: This does not depend on the mass!!

energy in simple harmonic motion

max

Energy in Simple Harmonic Motion
  • As expected Energy is conserved.
    • Oscillates between potential and kinetic energy.

All Potential

Energy

All Kinetic

Energy

min

energy for springs

Total Energy

Kinetic Energy

Potential Energy

Energy for springs.
  • As expected Energy is conserved.
    • Oscillates between potential and kinetic energy.

PE = ½kx(t)2 = ½kA2cos2(t)

KE = ½mv(t)2 = ½m2A2sin2(t)

KEmax = ½m2A2 = PEmax = ½kA2

Note: 2 = k/m

capa 6 10
CAPA 6-10
  • A 368g mass vibrates according to the equation x = 0.346 sin (5.20 t) where x is in meters and t is in seconds. Determine the amplitude.
  • Determine the frequency.
  • Determine the period.
  • Determine the total energy.
  • Determine the kinetic energy when x is 12.5 cm.

x = 0.346 sin (5.20 t)

Compare this with the standard formula:

x = A sin (t) A = 0.346 m and  = 5.20 rad/s

capa 6
CAPA 6
  • A 368g mass vibrates according to the equation x = 0.346 sin (5.20 t) where x is in meters and t is in seconds. Determine the amplitude.

x = 0.346 sin (5.20 t)

Compare this with the standard formula:

x = A sin (t) A = 0.346 m and  = 5.20 rad/s

Amplitude = 0.346 m

capa 7
CAPA 7

A 368g mass vibrates according to the equation x = 0.346 sin (5.20 t) where x is in meters and t is in seconds.

  • Determine the frequency.

x = 0.346 sin (5.20 t)

Compare this with the standard formula:

x = A sin (t) A = 0.346 m and  = 5.20 rad/s

2f =   f = /2 = 5.20/(2*3.14159) = 0.828 s-1

capa 8
CAPA 8

A 368g mass vibrates according to the equation x = 0.346 sin (5.20 t) where x is in meters and t is in seconds.

  • Determine the period.

T = 1/f  T = 1/0.828 s-1 = 1.21 s

capa 9
CAPA 9

A 368g mass vibrates according to the equation x = 0.346 sin (5.20 t) where x is in meters and t is in seconds.

  • Determine the total energy.

Total Energy = ½ kA2

2 = k/m  k = m2 = 0.368 * 5.22 = 9.95 N/m

Total Energy = ½ * 9.95 N/m * (.346 m)2

Total Energy = 0.596 J

capa 10
CAPA 10

A 368g mass vibrates according to the equation x = 0.346 sin (5.20 t) where x is in meters and t is in seconds.

  • Determine the kinetic energy when x is 12.5 cm

KE = Total Energy – Potential Energy

KE = .596 – ½kx2

KE = .596 – ½*9.95 *.1252

KE = .518 J

next time
Next Time
  • Quiz on Chapter 15
    • First Law
    • Engines and Efficiency.
  • Please see me with any questions or comments.

See you on Wednesday.