PHYSICS 231 INTRODUCTORY PHYSICS I Lecture 20 Last Lecture Heat Engine Refrigerator Heat Pump Qhot engine Qhot W fridge Qcold W Qcold W Tcold e eCarnot 1 Qhot Thot Qc h Tc h COP COPCarnot W Thot Tcold W Qhot Qcold Entropy Measure of Disorder of the system randomness ignorance Entropy S kBlog N N of possible arrangements for fixed E and Q Number of ways for 12 molecules to arrange themselves in two halves of container S is greater if molecules spread evenly in both halves 2nd Law of Thermodynamics version 2 The Total Entropy of the Universe can never decrease but entropy of system can increase or decrease On a macroscopic level one finds that adding heat raises entropy S Q T Temperature in Kelvin 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 S Q TB Q TA This can only occur if S 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 S Senvironment Qcold Qhot 0 Tcold Thot W Qcold Tcold e 1 1 eCarnot Qhot Qhot Thot Chapter 13 Vibrations and Waves Hooke s Law Reviewed F kx When x is positive F is negative When at equilibrium x 0 F 0 When x is negative F is positive Sinusoidal Oscillation If we extend the mass and let go the pen traces a sine wave Graphing x vs t A T A amplitude length m T period time s Period and Frequency A T x Acos t T 2 Amplitude A Period T 2 T Frequency f 1 T Angular frequency f 2 Phases Often a phase the peak is included to shift the timing of x Acos t Acos t t 0 for peak at t t0 Phase of 90 degrees changes cosine to sine cos t sin t 2 Velocity and Acceleration vs time 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 x T v T a T v and a vs t x A cos t v vmax sin t a amax cos t Find vmax with E conservation 1 2 1 2 kA mvmax 2 2 k vmax A m Find amax using F ma kx ma kA cos t mamax cos t k amax A m Connection to Circular Motion Projection on axis circular motion with constant angular velocity Simple Harmonic Motion What is Circular motion Angular speed Radius A Simple Harmonic Motion Cons of E v max Speed v A k A m k m k m Formula Summary 1 f T 2 2 f T x A cos t v Asin t a 2 A cos t k m 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 4a An object undergoing simple harmonic motion follows the expression x t 4 2 cos t 3 Where x will be in cm if t is in seconds The amplitude of the motion is a 1 cm b 2 cm c 3 cm d 4 cm e 4 cm Example 13 4b An object undergoing simple harmonic motion follows the expression x t 4 2 cos t 3 Here x will be in cm if t is in seconds The period of the motion is a 1 3 s b 1 2 s c 1 s d 2 s e 2 s Example 13 4c An object undergoing simple harmonic motion follows the expression x t 4 2 cos t 3 Here x will be in cm if t is in seconds The frequency of the motion is a 1 3 Hz b 1 2 Hz c 1 Hz d 2 Hz e Hz Example 13 4d An object undergoing simple harmonic motion follows the expression x t 4 2 cos t 3 Here x will be in cm if t is in seconds The angular frequency of the motion is a 1 3 rad s b 1 2 rad s c 1 rad s d 2 rad s e rad s Example 13 4e An object undergoing simple harmonic motion follows the expression x t 4 2 cos t 3 Here x will be in cm if t is in seconds The object will pass through the equilibrium position at the times t seconds a b c d e 2 1 0 1 2 1 5 0 5 0 5 1 5 2 5 1 5 1 0 5 0 0 5 1 0 1 5 4 2 0 2 4 2 5 0 5 1 5 3 5
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