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Midterm Exam 2 March 29, 7:30-8:30 pm, STEW 183 will cover Chapters 6-10. Lecture 18 Purdue University, Physics 220 1 PHYSICS 220 Lecture 18 Elasticity and Oscillations Lecture 18 Purdue University, Physics 220 2 Contact Force: Spring Force exerted by a spring is directly proportional to the amount by which it is stretched or compressed. Fspring = - k x always trying to restore its original length Lecture 4 Purdue University, Physics 220 3 Hooke’s Law Revisited F = - k DL; where k depends on size of the bar and material Hooke’s Law F/A = Y *DL/L0 k = Y A/L0 • Y (Young’s Modules) independent of L or A Has the unit of (N/m2) pressure (i.e., pascal) Inherent stiffness of a material • For many materials, Y is the same for compression and tension, elastic constant • Stress = Y*Strain Define Strain = DL / L0 Stress = F/A Lecture 18 Purdue University, Physics 220 4 Beyond Hook’s Law Deformed beyond elastic limit material will no longer return to its original length even when F is completely removed. Lecture 18 Purdue University, Physics 220 5 Simple Harmonic Oscillation The force exerted by a spring is proportional to the distance the spring is stretched or compressed from its relaxed position. FX = -k x x is the displacement from the relaxed position and k is the constant of proportionality. Lecture 18 Purdue University, Physics 220 6 Potential Energy in Spring • Force of spring is conservative F = -k x DPE = -W W = -½ k x2 DPE= ½ k x2 Force work x Work done only depends on initial and final position Lecture 18 Purdue University, Physics 220 7 Energy Conservation • A mass is attached to a spring and set to motion. The maximum displacement is x=A SWnc = DKE + DPE 0 = DKE + DPE or Energy PE+KE is constant! Energy = ½ k x2 + ½ m v2 PE At maximum displacement x=A, v = 0 S Energy = ½ k A2 + 0 At zero displacement x = 0 Energy = 0 + ½ mvm2 0 Since Total Energy is same ½ k A2 = ½ m vm2 m vm = sqrt(k/m) A x=0 Lecture 18 Purdue University, Physics 220 x x 8 Vertical Mass and Spring • If we include gravity, there are two forces acting on mass. With mass, new equilibrium position has spring stretched d At equilibrium: F=kd–mg=0 d = mg/k Let this point be y=0 At a displacement, F=k(d-y)–mg=-k y Total energy: E=1/2k(d-y)2+mgy+1/2mv2 =1/2kd2+1/2ky2+1/2mv2 Same as the horizontal case, except for a new equilibrium position Lecture 18 Purdue University, Physics 220 9 What is harmonic motion? Lecture 18 Purdue University, Physics 220 10 iClicker A mass on a spring oscillates back & forth with simple harmonic motion of amplitude A. A plot of displacement (y) versus time (t) is shown below. At what points during its oscillation is the total energy of the mass and spring a maximum? (ignore gravity). A) When y = +A or -A (i.e. maximum displacement) B) When y = 0 (i.e. zero displacement) C) The energy of the system is constant “When the kinetic energy is at a minimum, the potential energy is at a maximum and vice-versa.” y +A t Lecture 18 -A Purdue University, Physics 220 11 Exercise A mass on a spring oscillates back & forth with simple harmonic motion of amplitude A. A plot of displacement (y) versus time (t) is shown below. At what points during its oscillation does the speed of the block reach maximum? A) When y = +A or -A (i.e. maximum displacement) B) When y = 0 (i.e. zero displacement) C) The speed of the mass is constant y +A t -A Lecture 18 Purdue University, Physics 220 12 Question • A spring oscillates back and forth on a frictionless horizontal surface. A camera takes pictures of the position every 1/10th of a second. Which plot best shows the positions of the mass. A) EndPoint Equilibrium EndPoint B) EndPoint CORRECT Equilibrium EndPoint Equilibrium EndPoint C) EndPoint Lecture 18 Purdue University, Physics 220 13 Analogy to Circular Motion What does moving in a circle have to do with moving back & forth in a straight line? x = A cos = A cos (wt) since = w t x x 1 1 2 A 3 A 8 2 8 y 7 4 6 0 2 -A 5 Lecture 18 7 3 4 3 2 6 5 Purdue University, Physics 220 14 Analogy to Circular Motion Period = T (seconds per cycle) Frequency = f = 1/T (cycles per second) Angular frequency = w = 2f = 2/T Lecture 18 Purdue University, Physics 220 15 Analogy to Circular Motion x(t) = Acos(wt) x(t) = Asin(wt) V(t) = -Awsin(wt) V(t) = Awcos(wt) a(t) = -Aw2cos(wt) =-w2 x(t) a(t) = -Aw2sin(wt) =-w2 x(t) For spring: F = ma = -kx, so a(t) = -(k/m)x(t). By analogy, we have w2 = k/m xmax = A vmax = Aw amax = Aw2 Lecture 18 Purdue University, Physics 220 16 Lecture 18 Purdue University, Physics 220 17 Example A 3 kg mass is attached to a spring (k=24 N/m). It is stretched 5 cm. At time t=0 it is released and oscillates. Which equation describes the position as a function of time x(t) (in cm) = A) 5 sin(wt) B) 5 cos(wt) C) 24 sin(wt) D) 24 cos(wt) E) -24 cos(wt) We are told at t=0, x = +5 cm. x(t) = 5 cos(wt) only one that works. Lecture 18 Purdue University, Physics 220 18 Example A 3 kg mass is attached to a spring (k=24 N/m). It is stretched 5 cm. At time t=0 it is released and oscillates. What is the total energy of the block spring system? E = PE + KE At t=0, x = 5 cm and v=0: E = ½ k x2 + 0 = ½ (24 N/m) (5 cm)2 = 0.03 J Lecture 18 Purdue University, Physics 220 19 Example A 3 kg mass is attached to a spring (k=24 N/m). It is stretched 5 cm. At time t=0 it is released and oscillates. What is the maximum speed of the block? E = PE + KE When x = 0, maximum speed: E = ½ m v2 + 0 (0.03 J) = ½ 3 (kg) v2 J=kg*m2/s2 v = .14 m/s Lecture 18 Purdue University, Physics 220 20 Example A 3 kg mass is attached to a spring (k=24 N/m). It is stretched 5 cm. At time t=0 it is released and oscillates. How long does it take for the block to return to x=+5cm? w = sqrt(k/m) = sqrt(24/3) = 2.83 radians/sec Returns to original position after 2 radians T = 2 w = 6.28 / 2.83 = 2.2 seconds Lecture 18 Purdue University, Physics 220 21 Lecture 18 Purdue University, Physics 220 22 Pendulum Motion • For small angles T = mg Tx = -mg (x/L) Note: F proportional to x! Fx = m ax -mg (x/L) = m ax ax = -(g/L) x Recall for SHO a = -w2 x L w = sqrt(g/L) T = 2 sqrt(L/g) T Period does not depend on A, or m! x m mg Lecture 18 Purdue University, Physics 220 23 Question Suppose a grandfather clock (a simple pendulum) runs slow. In order to make it run on time you should: CORRECT A) Make the pendulum shorter B) Make the pendulum longer w g L 2 L T 2 w g Lecture 18 Purdue University, Physics 220 24 iClicker In Case 1 a mass on a spring oscillates back and forth. In Case 2 the mass is doubled but the spring and the amplitude of the oscillation is the same as in Case 1. Which case has the smallest maximum velocity? A) Case 1 B) Case 2 C) Same Same maximum Kinetic Energy K = ½ m v2 Lecture 18 smaller mass requires larger v Purdue University, Physics 220 25