2Last Lecture Heat Engine Refrigerator, Heat Pump Qhot Qhot W W engine QcoldWQhotfridgeQcoldW
3EntropyMeasure of Disorder of the system (randomness, ignorance)Entropy: S = kBlog(N) N = # of possible arrangements for fixed E and QNumber of ways for 12 molecules to arrange themselves in two halves of container.S is greater if molecules spread evenly in both halves.
42nd Law of Thermodynamics (version 2) The Total Entropy of the Universe can neverdecrease.(but entropy of system can increase or decrease)On a macroscopic level, one finds that adding heat raises entropy:Temperature in Kelvin!
5Why does Q flow from hot to cold? Consider two systems, one with TA and one with TBAllow Q > 0 to flow from TA to TBEntropy changes by: DS = Q/TB - Q/TAThis can only occur if DS > 0, requiring TA > TB.System will achieve more randomness by exchanging heat until TB = TA
6Carnot EngineCarnot cycle is most efficient possible, because the total entropy change is zero.It is a “reversible process”.For real engines:
8Hooke’s Law Reviewed When x is positive , F is negative ; When at equilibrium (x=0), F = 0 ;When x is negative , F is positive ;
9Sinusoidal Oscillation If we extend the mass, and let go, the pen traces a sine wave.
10Graphing x vs. tA : amplitude (length, m) T : period (time, s)AT
11Period and Frequency Amplitude: A Period: T Frequency: f = 1/T Angular frequency:
12Phases Often a phase is included to shift the timing of the peak: for peak atPhase of 90-degrees changes cosine to sine
13Velocity and Acceleration vs. time xvTVelocity and Acceleration vs. timeVelocity 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) xTT
14v and a vs. tFind vmax with E conservationFind amax using F=ma
15Connection to Circular Motion Projection on axiscircular motion with constant angular velocity Simple Harmonic Motion
16What is w? Simple Harmonic Motion Circular motion Cons. of E: Angular speed: Radius: A=> Speed: v=A
18Example13.1 a) 0.153 J b) 0.783 m/s c) 17.5 m/s2 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) Jb) m/sc) 17.5 m/s2
19Example 13.2A 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) cmb) cm/s
20Example 13.3A 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) cm
21Example 13.4aAn object undergoing simple harmonic motion follows the expression,Where x will be in cm if t is in secondsThe amplitude of the motion is:a) 1 cmb) 2 cmc) 3 cmd) 4 cme) -4 cm
22Example 13.4bAn object undergoing simple harmonic motion follows the expression,Here, x will be in cm if t is in secondsThe period of the motion is:a) 1/3 sb) 1/2 sc) 1 sd) 2 se) 2/ s
23Example 13.4cAn object undergoing simple harmonic motion follows the expression,Here, x will be in cm if t is in secondsThe frequency of the motion is:a) 1/3 Hzb) 1/2 Hzc) 1 Hzd) 2 Hze) Hz
24Example 13.4dAn object undergoing simple harmonic motion follows the expression,Here, x will be in cm if t is in secondsThe angular frequency of the motion is:a) 1/3 rad/sb) 1/2 rad/sc) 1 rad/sd) 2 rad/se) rad/s
25Example 13.4eAn object undergoing simple harmonic motion follows the expression,Here, x will be in cm if t is in secondsThe object will pass through the equilibrium position at the times, t = _____ secondsa) …, -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,