Lecture 21: Ideal Spring and Simple Harmonic Motion z New Material: Textbook Chapters 10.1, 10.2 and 10.3 Physics 101: Lecture 21, Pg 1 Ideal Springs z 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). ÎFX = -k x Where 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 FX = 0 x x=0 Physics 101: Lecture 21, Pg 2 Ideal Springs z Hooke’s Law: The force exerted by a spring is proportional to the distance the spring is stretched or compressed from its relaxed position. ÎFX = -k x Where x is the displacement from the relaxed position and k is the constant of proportionality. (often called “spring constant”) relaxed position FX = -kx > 0 x x<0 x=0 Physics 101: Lecture 21, Pg 3 Ideal Springs z Hooke’s Law: The force exerted by a spring is proportional to the distance the spring is stretched or compressed from its relaxed position. ÎFX = -k x Where x is the displacement from the relaxed position and k is the constant of proportionality. (often called “spring constant”) relaxed position FX = - kx < 0 x>0 x x=0 Physics 101: Lecture 21, Pg 4 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 5 What does moving along a circular path have to do with moving back & forth in a straight line (oscillation about equilibrium) ?? x = R cos θ = R cos (ωt) since θ = ω t x x 1 2 y R 3 R 8 θ 7 4 6 5 0 -R 1 2 8 7 3 π 2 π 4 3π 2 θ 6 5 Physics 101: Lecture 21, Pg 6 Velocity and Acceleration z Using again the reference circle one finds for the velocity v = - vT sin θ = - A ω sin (ω t) and for the acceleration a = - ac cos θ = - A ω2 cos (ω t) with ω in [rad/s] Physics 101: Lecture 21, Pg 7 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) CORRECT 3. The speed of the mass is constant +A x t -A 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 magnitude of the acceleration of the block biggest? 1. When x = +A or -A (i.e. maximum displacement) CORRECT 2. When x = 0 (i.e. zero displacement) 3. The acceleration of the mass is constant +A x t -A Physics 101: Lecture 21, Pg 9 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 Physics 101: Lecture 21, Pg 10 Simple Harmonic Motion: At t=0 s, x=A or x(t) = [A]cos(ωt) v(t) = -[Aω]sin(ωt) a(t) = -[Aω2]cos(ωt) At t=0 s, x=0 m x(t) = [A]sin(ωt) OR v(t) = [Aω]cos(ωt) a(t) = -[Aω2]sin(ωt) xmax = A Period = T (seconds per cycle) vmax = Aω Frequency = f = 1/T (cycles per second) amax = Aω2 Angular frequency = ω = 2πf = 2π/T For spring: ω2 = k/m Physics 101: Lecture 21, Pg 11 Elastic Potential Energy z Work done by the (average) restoring force of the spring is W = |Fave| s cos θ = ½ k ( x0+xf) (x0-xf) = = ½ k (x02 – xf2) = Epot,elastic,0- Epot,elastic,f The elastic potential energy Epot,elastic = ½ k x2 has to be considered in addition to kinetic and gravitational potential energy when calculating the total mechanical energy of an object. Physics 101: Lecture 21, Pg 12