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Physics 101: Lecture 18 Elasticity and Oscillations

Exam III. Physics 101: Lecture 18 Elasticity and Oscillations. Example: hose.

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Physics 101: Lecture 18 Elasticity and Oscillations

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  1. Exam III Physics 101: Lecture 18 Elasticity and Oscillations

  2. Example: hose A garden hose w/ inner diameter 2 cm, carries water at 2.0 m/s. To spray your friend, you place your thumb over the nozzle giving an effective opening diameter of 0.5 cm. What is the speed of the water exiting the hose? What is the pressure difference between inside the hose and outside? Continuity Equation Bernoulli Equation

  3. Example: Lift a House Calculate the net lift on a 15 m x 15 m house when a 30 m/s wind (1.29 kg/m3) blows over the top. 48

  4. Overview • Springs (review) • Restoring force proportional to displacement • F = -k x • U = ½ k x2 • Today • Young’s Modulus • Simple Harmonic Motion • Springs Revisited 05

  5. Simple Harmonic Motion • Vibrations • Vocal cords when singing/speaking • String/rubber band • Simple Harmonic Motion • Restoring force proportional to displacement • Springs F = -kx 11

  6. x +A t -A Question A mass on a spring oscillates back & forth with simple harmonic motion of amplitude A. A plot of displacement (x) versus time (t) is shown below. At what points during its oscillation is the magnitude of the acceleration of the block biggest? 1. When x = +A or -A (i.e. maximum displacement) 2. When x = 0 (i.e. zero displacement) 3. The acceleration of the mass is constant 17

  7. Potential Energy in Spring • Force of spring is Conservative • F = -k x • W = -1/2 k x2 • Work done only depends on initial and final position • Define Potential Energy PEspring = ½ k x2 Force work x 20

  8. m x x=0 PES x 0 ***Energy in SHM*** • A mass is attached to a spring and set to motion. The maximum displacement is x=A • Energy = PE + KE = constant! = ½ k x2 + ½ m v2 • At maximum displacement x=A, v = 0 Energy = ½ k A2 + 0 • At zero displacement x = 0 Energy = 0 + ½ mvm2 Since Total Energy is same ½ k A2 = ½ m vm2 vm = sqrt(k/m) A 25

  9. x +A t -A Question A mass on a spring oscillates back & forth with simple harmonic motion of amplitude A. A plot of displacement (x) versus time (t) is shown below. At what points during its oscillation is the speed of the block biggest? 1. When x = +A or -A (i.e. maximum displacement) 2. When x = 0 (i.e. zero displacement) 3. The speed of the mass is constant 29

  10. 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. 1 2 3 EndPoint Equilibrium EndPoint EndPoint Equilibrium EndPoint EndPoint Equilibrium EndPoint 38

  11. Springs and Simple Harmonic Motion X=0 X=A; v=0; a=-amax X=0; v=-vmax; a=0 X=-A; v=0; a=amax X=0; v=vmax; a=0 X=A; v=0; a=-amax X=-A X=A 32

  12. x = R cos q =R cos (wt) sinceq = wt x 1 1 R 2 2 3 3  0 y 4 6 -R 4 6 5 5 What does moving in a circlehave to do with moving back & forth in a straight line ?? x 8 8 q R 7 7 34

  13. SHM and Circles

  14. Simple Harmonic Motion: x(t) = [A]cos(t) v(t) = -[A]sin(t) a(t) = -[A2]cos(t) x(t) = [A]sin(t) v(t) = [A]cos(t) a(t) = -[A2]sin(t) OR Period = T (seconds per cycle) Frequency = f = 1/T (cycles per second) Angular frequency =  = 2f = 2/T For spring: 2 = k/m xmax = A vmax = A amax = A2 36

  15. 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) = A) 5 sin(wt) B) 5 cos(wt) C) 24 sin(wt) D) 24 cos(wt) E) -24 cos(wt) 39

  16. 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? 43

  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. What is the maximum speed of the block? 46

  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. How long does it take for the block to return to x=+5cm? 49

  19. Summary • Springs • F = -kx • U = ½ k x2 • w = sqrt(k/m) • Simple Harmonic Motion • Occurs when have linear restoring force F= -kx • x(t) = [A] cos(wt) or [A] sin(wt) • v(t) = -[Aw] sin(wt) or [Aw] cos(wt) • a(t) = -[Aw2] cos(wt) or -[Aw2] sin(wt) 50

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