Physics 231 Lecture 32 Main points of last lecture Carnot cycle and Carnot engine Entropy S Qreversible T Entropy is the log of the probability of a state The total entropy of all interacting systems can only increase or stay the same during a process It can never decrease Hooke s law 1 Fs kx PEs kx 2 2 Main points of today s lecture Springs and masses Simple harmonic motion of a spring x A cos t 0 v x A sin t 0 a x 2 A cos t 0 2 k m 0 and A are constants Pendulum L T 2 g max cos 2 ft 0 Example In a room that is 2 44 m high a spring unstrained length 0 30 m hangs from the ceiling From this spring a board hangs so that its 1 98 m length is perpendicular to the floor the lower end just extending to but not touching the floor The board weighs 102 N What is the spring constant of the spring 0 Ftotal ky mg k d mg kd mg k d 2 44 1 98 0 3 m mg 102 N 638 N m 2 44 1 98 0 3 m d Conceptual question An object can oscillate around a any equilibrium point b any stable equilibrium point c certain stable equilibrium points d any point provided the forces exerted on it obey Hooke s law e any point Example A 3 0 kg block is placed between two horizontal springs Neither spring is strained when the block is located at the position labeled x 0 in the drawing What is the effective Hook s law constant keff which describes the restoring force Fs k1 x k 2 x k1 k 2 x keff k1 k 2 keff 1100 N M Quiz The block is then displaced a distance of 0 070 m from the position where x 0 and released from rest What is the speed of the block when it passes the x 0 position Hint use conservation of energy 1 1 KE f PE0 mv 2f keff x 2 a 2 55 m s 2 2 b 1 98 m s keff 2 keff 2 vf x vf x c 1 47 m s m m d 1 34 m s 1100 0 07m 1 34m s vf 3 Stress and strain for springs If you cut a spring in half the restoring force of each piece will double If you cut it into thirds the restoring forces will triple In this sense the Hook s constant depends on length similar the tensile strength of solids AY F L F L Y A L0 L0 If we shorten a spring with original length L0 and spring constant k L0 k L k L 0 L0 L Analogy to circular motion The shadow of a ball on rotating turntable moves in simple harmonic motion just as a mass on a spring This is because the horizontal motion for both satisfy F kx max ax k m x For the circular motion r A v A a 2A t 0 Thus x r cos A cos t 0 vx v sin Asin t 0 ax acos 2Acos t 0 2x Thus if we choose 2 k m for the circular motion the shadow of the circling ball will have the same acceleration initial position and initial velocity and with therefore track the simple harmonic motion of the mass and spring exactly We therefore can determine the dependence of position velocity and acceleration on time for the mass and spring from the circular motion vx v a v v A ax x Example A 100 coil spring has a spring constant of 430 N m It is cut into two shorter springs each of which has 50 coils One end of a 50 coil spring is attached to a wall An object of mass 45 kg is attached to the other end of the spring and the system is set in horizontal oscillation What is the angular frequency of the motion k50 2k100 2 430 N m 860 N m keff m 860 N m 4 37 Hz 45kg If mass spring system was arranged vertically with the mass suspended from the 50 coil spring how would the frequency change a it would be smaller because gravity subtracts from the spring force at the bottom of its motion b it would be larger because gravity adds to the spring force at the top of its motion c it would be exactly the same Gravity only displaces the equilibrium point so that the equilibrium length is greater
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