Physics 101: Lecture 21, Pg 1 Lecture 21: Ideal Spring and Simple Harmonic Motion l New Material: Textbook Chapters 10.1, 10.2 and 10.3

Physics 101: Lecture 21, Pg 2 Ideal Springs l Hooke’s Law: l Hooke’s Law: The force exerted by a spring is proportional to the distance the spring is stretched or compressed from its relaxed position (for small x). è F X = -k xWhere x is the displacement from the relaxed position and k is the constant of proportionality. (often called “spring constant”) SI unit of k: [N/m] relaxed position F X = 0 x x=0

Physics 101: Lecture 21, Pg 3 Ideal Springs Ideal Springs l Hooke’s Law: l Hooke’s Law: The force exerted by a spring is proportional to the distance the spring is stretched or compressed from its relaxed position. è F X = -k xWhere x is the displacement from the relaxed position and k is the constant of proportionality. (often called “spring constant”) relaxed position F X = -kx > 0 x x  0 x=0

Physics 101: Lecture 21, Pg 4 Ideal Springs l Hooke’s Law: l Hooke’s Law: The force exerted by a spring is proportional to the distance the spring is stretched or compressed from its relaxed position. è F X = -k xWhere x is the displacement from the relaxed position and k is the constant of proportionality. (often called “spring constant”) F X = - kx < 0 x x > 0 relaxed position x=0

Physics 101: Lecture 21, Pg 5 Simple Harmonic Motion Consider the friction-free motion of an object attached to an ideal spring, i.e. a spring that behaves according to Hooke’s law. How does displacement, velocity and acceleration of the object vary with time ? Analogy: Simple harmonic motion along x x component of uniform circular motion

Physics 101: Lecture 21, Pg 6 What does moving along a circular path have to do with moving back & forth in a straight line (oscillation about equilibrium) ?? y x -R R 0  R  x x = R cos  = R cos (  t ) since  =  t

Physics 101: Lecture 21, Pg 7 Velocity and Acceleration l Using again the reference circle one finds for the velocity v = - v T sin  = - A  sin (  t) and for the acceleration a = - a c cos  = - A  2 cos (  t) with  in [rad/s]

Physics 101: Lecture 21, Pg 8 Concept 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 CORRECT

Physics 101: Lecture 21, Pg 9 Concept 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 CORRECT

Physics 101: Lecture 21, Pg 10 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

Physics 101: Lecture 21, Pg 11 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 At t=0 s, x=A or At t=0 s, x=0 m

Physics 101: Lecture 21, Pg 12 Elastic Potential Energy l Work done by the (average) restoring force of the spring is W = |F ave | s cos  = ½ k ( x 0 +x f ) (x 0 -x f ) = = ½ k (x 0 2 – x f 2 ) = E pot,elastic, 0 - E pot,elastic, f The elastic potential energy E pot,elastic = ½ k x 2 has to be considered in addition to kinetic and gravitational potential energy when calculating the total mechanical energy of an object.