2 Last Lecture Heat Engine Refrigerator, Heat Pump Qhot Qhot W W engine QcoldWQhotfridgeQcoldW
3 EntropyMeasure 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.
4 2nd 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!
5 Why 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
6 Carnot EngineCarnot cycle is most efficient possible, because the total entropy change is zero.It is a “reversible process”.For real engines:
8 Hooke’s Law Reviewed When x is positive , F is negative ; When at equilibrium (x=0), F = 0 ;When x is negative , F is positive ;
9 Sinusoidal Oscillation If we extend the mass, and let go, the pen traces a sine wave.
10 Graphing x vs. tA : amplitude (length, m) T : period (time, s)AT
11 Period and Frequency Amplitude: A Period: T Frequency: f = 1/T Angular frequency:
12 Phases Often a phase is included to shift the timing of the peak: for peak atPhase of 90-degrees changes cosine to sine
13 Velocity 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
14 v and a vs. tFind vmax with E conservationFind amax using F=ma
15 Connection to Circular Motion Projection on axiscircular motion with constant angular velocity Simple Harmonic Motion
16 What is w? Simple Harmonic Motion Circular motion Cons. of E: Angular speed: Radius: A=> Speed: v=A
18 Example13.1 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
19 Example 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
20 Example 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
21 Example 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
22 Example 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
23 Example 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
24 Example 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
25 Example 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,