# Nov. 26, 2001 Dr. Larry Dennis, FSU Department of Physics1 Physics 2053C – Fall 2001 Chapter 11 Simple Harmonic Motion.

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Nov. 26, 2001 Dr. Larry Dennis, FSU Department of Physics1 Physics 2053C – Fall 2001 Chapter 11 Simple Harmonic Motion

2 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 ) Engine THTH TLTL QHQH QLQL W

3 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? Engine THTH TLTL QHQH QLQL W

4 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 Q H – Q L =  U + W Q L = Q H – W = 200 – 75 J = 125 J Engine THTH TLTL QHQH QLQL W

5 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 – T L /T H = 0.375 T H – T L = 0.375 * T H T L = T H – 0.375 * T H T L = 0.625*T H T H = T L /0.625 = 190 / 0.625 = 304 K Engine THTH TLTL QHQH QLQL W

6 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. Engine THTH TLTL QHQH QLQL W

7 Periodic Motion Many kinds of motion are periodic. Examine within the context of: Forces Energy Equilibrium Position & Restoring Forces

8 Position vs. Time x = A sin(2  *t/T) Position = Amplitude * sin( 2  * time/Period) Amplitude = A Period = T t x

9 Spring Oscillations m k x x(t) = Acos(  t) or Asin(  t)  = 2  f = 2  /T =  (k/m)

10 Pendulum Oscillations  (t) = A cos(  t) or  (t) = A sin(  t)  = 2  f = 2  /T =  (g/L) L Note: This does not depend on the mass!! T mg Net Force

11 Energy in Simple Harmonic Motion As expected Energy is conserved. Oscillates between potential and kinetic energy.  max All Kinetic Energy  min All Potential Energy

12 Energy for springs. As expected Energy is conserved. Oscillates between potential and kinetic energy. PE = ½kx(t) 2 = ½kA 2 cos 2 (  t) KE = ½mv(t) 2 = ½m  2 A 2 sin 2 (  t) KE max = ½m  2 A 2 = PE max = ½kA 2 Note:  2 = k/m Total Energy Potential Energy Kinetic Energy

13 CAPA 6-10 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. 7.Determine the frequency. 8.Determine the period. 9.Determine the total energy. 10.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

14 CAPA 6 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

15 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. 7.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

16 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. 8.Determine the period. T = 1/f  T = 1/0.828 s -1 = 1.21 s

17 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. 9.Determine the total energy. Total Energy = ½ kA 2  2 = k/m  k = m  2 = 0.368 * 5.2 2 = 9.95 N/m Total Energy = ½ * 9.95 N/m * (.346 m) 2 Total Energy = 0.596 J

18 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. 10.Determine the kinetic energy when x is 12.5 cm KE = Total Energy – Potential Energy KE =.596 – ½kx 2 KE =.596 – ½*9.95 *.125 2 KE =.518 J

19 Next Time Quiz on Chapter 15 First Law Engines and Efficiency. Please see me with any questions or comments. See you on Wednesday.

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