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Lecture 21: Ideal Spring and Simple Harmonic Motion. New Material: Textbook Chapters 10.1, 10.2 and 10.3 . relaxed position. F X = 0. x. x=0. Ideal Springs.
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Lecture 21: Ideal Spring and Simple Harmonic Motion • New Material: Textbook Chapters 10.1, 10.2 and 10.3
relaxed position FX = 0 x x=0 Ideal Springs • 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). • FX = -k xWhere xis the displacement from the relaxed position and k is the constant of proportionality. (often called “spring constant”)
Ideal Springs • Hooke’s Law:The force exerted by a spring is proportional to the distance the spring is stretched or compressed from its relaxed position. • FX = -k xWhere xis 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
Ideal Springs • Hooke’s Law:The force exerted by a spring is proportional to the distance the spring is stretched or compressed from its relaxed position. • FX = -k xWhere xis 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
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
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 along a circular path have to do with moving back & forth in a straight line (oscillation about equilibrium) ?? x 8 8 q R 7 7
Velocity and Acceleration • Using again the reference circle one finds for the velocity v = - vT sin q = - A w sin (w t) and for the acceleration a = - ac cos q = - A w2 cos (w t) with w in [rad/s]
x +A CORRECT t -A 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
x +A CORRECT t -A 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
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
Simple Harmonic Motion: At t=0 s, x=A or At t=0 s, x=0 m x(t) = [A]cos(t) v(t) = -[A]sin(t) a(t) = -[A2]cos(t) x(t) = [A]sin(t) v(t) = [A]cos(t) a(t) = -[A2]sin(t) OR Period = T (seconds per cycle) Frequency = f = 1/T (cycles per second) Angular frequency = = 2f = 2/T For spring: 2 = k/m xmax = A vmax = A amax = A2
Elastic Potential Energy • Work done by the (average) restoring force of the spring is W = |Fave| s cos q = ½ k ( x0+xf) (x0-xf) = = ½ k (x02 – xf2) = Epot,elastic,0- Epot,elastic,f The elastic potential energy has to be considered when Calculating the total mechanical energy of an object attached to a spring.
