Physics 101: Lecture 18, Pg 1 Physics 101: Lecture 18 Elasticity and Oscillations Exam III

Physics 101: Lecture 18, Pg 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? Bernoulli Equation Continuity Equation

Physics 101: Lecture 18, Pg 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/m 3 ) blows over the top. 48

Physics 101: Lecture 18, Pg 4 Overview l Springs (review) è Restoring force proportional to displacement è F = -k x è U = ½ k x 2 l Today è Young’s Modulus è Simple Harmonic Motion è Springs Revisited 05

Physics 101: Lecture 18, Pg 5 Simple Harmonic Motion l Vibrations è Vocal cords when singing/speaking è String/rubber band l Simple Harmonic Motion è Restoring force proportional to displacement è Springs F = -kx 11

Physics 101: Lecture 18, Pg 6 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 +A t -A x 17

Physics 101: Lecture 18, Pg 7 Potential Energy in Spring l Force of spring is Conservative è F = -k x è W = -1/2 k x 2 è Work done only depends on initial and final position è Define Potential Energy PE spring = ½ k x 2 Force x work 20

Physics 101: Lecture 18, Pg 8 ***Energy in SHM*** l A mass is attached to a spring and set to motion. The maximum displacement is x=A è Energy = PE + KE = constant! = ½ k x 2 + ½ m v 2 è At maximum displacement x=A, v = 0 Energy = ½ k A è At zero displacement x = 0 Energy = 0 + ½ mv m 2 Since Total Energy is same ½ k A 2 = ½ m v m 2 v m = sqrt(k/m) A m x x=0 0 x PE S 25

Physics 101: Lecture 18, Pg 9 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 +A t -A x 29

Physics 101: Lecture 18, Pg 10Question l A spring oscillates back and forth on a frictionless horizontal surface. A camera takes pictures of the position every 1/10 th of a second. Which plot best shows the positions of the mass. EndPoint Equilibrium EndPoint Equilibrium EndPoint Equilibrium

Physics 101: Lecture 18, Pg 11 X=0 X=A X=-A X=A; v=0; a=-a max X=0; v=-v max ; a=0X=-A; v=0; a=a max X=0; v=v max ; a=0 X=A; v=0; a=-a max Springs and Simple Harmonic Motion 32

Physics 101: Lecture 18, Pg 12 What does moving in a circle have to do with moving back & forth in a straight line ?? y x -R R 0  R  x x = R cos  = R cos (  t ) since  =  t 34

Physics 101: Lecture 18, Pg 13 SHM and Circles

Physics 101: Lecture 18, Pg 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) x max = A v max = A  a max = A  2 Period = T (seconds per cycle) Frequency = f = 1/T (cycles per second) Angular frequency =  = 2  f = 2  /T For spring:  2 = k/m OR 36

Physics 101: Lecture 18, Pg 15Example 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(  t)B) 5 cos(  t)C) 24 sin(  t) D) 24 cos(  t)E) -24 cos(  t) 39

Physics 101: Lecture 18, Pg 16Example 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

Physics 101: Lecture 18, Pg 17Example 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

Physics 101: Lecture 18, Pg 18Example 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

Physics 101: Lecture 18, Pg 19Summary l Springs è F = -kx è U = ½ k x 2   = sqrt(k/m) l Simple Harmonic Motion è Occurs when have linear restoring force F= -kx  x(t) = [A] cos(  t) or [A] sin(  t)  v(t) = -[A  ] sin(  t) or [A  ] cos(  t)  a(t) = -[A   ] cos(  t) or -[A   ] sin(  t) 50