Lecture 18: Elasticity and Oscillations I l Simple Harmonic Motion: Definition l Springs: Forces l Springs: Energy l Simple Harmonic Motion: Equations

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

Springs: Forces 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. relaxed position (equilibrium position) F X = 0 +x x=0

Springs: Forces l Negative displacement: è Positive force è Positive acceleration l Zero displacement (at equilibrium): è Zero force è Zero acceleration l Positive displacement: è Negative force è Negative acceleration +x

Springs: Energy l Force of spring is Conservative: è F = -k xW = ½ k x 2 è Work done only depends on initial and final position. è Potential Energy: U spring = ½ k x 2 l A mass oscillating on a spring: è Energy = U + K = constant ½ k x 2 + ½ m v 2 = total energy m x x=0 0 x U

l Maximum Negative displacement: è Maximum potential energy è Zero kinetic energy l Zero displacement (at equilibrium): è Zero potential energy è Maximum kinetic energy l Maximum Positive displacement: è Maximum potential energy è Zero kinetic energy +x Springs: Energy

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

Spring Example l A mass of m = 4 kg is attached to a spring with k = 16 N/m and oscillates with an amplitude A = 0.15 m. What is the magnitude of the acceleration of the mass when the potential energy stored in the spring is equal to the kinetic energy of the mass? è First, use energy to find x where U = K: »U + K = constant è Second, use Hooke’s Law to find force on the mass: »F = -k x è Third, use Newton’s Second Law to find acceleration of mass: »F = m a

Spring Example l A mass of m = 4 kg is attached to a spring with k = 16 N/m and oscillates with an amplitude A = 0.15 m. What is the magnitude of the acceleration of the mass when the potential energy stored in the spring is equal to the kinetic energy of the mass? è First, use energy to find x where U = K: »U + K = constant »U + K = ½ k A 2 (since total energy is ½ k A 2 ) »U + U = ½ k A 2 (since K = U when they are equal) »k x 2 = ½ k A 2 (since U = ½ k x 2 when they are equal) »x 2 = ½ A 2 (simplify) = m

Spring Example l A mass of m = 4 kg is attached to a spring with k = 16 N/m and oscillates with an amplitude A = 0.15 m. What is the magnitude of the acceleration of the mass when the potential energy stored in the spring is equal to the kinetic energy of the mass? è Second, use Hooke’s Law to find force on the mass: »F = k x(we are only concerned with magnitude) = 1.70 N

Spring Example l A mass of m = 4 kg is attached to a spring with k = 16 N/m and oscillates with an amplitude A = 0.15 m. What is the magnitude of the acceleration of the mass when the potential energy stored in the spring is equal to the kinetic energy of the mass? è Third, use Newton’s Second Law to find acceleration of mass: »F = m a(we are only concerned with magnitude) = m/s 2

Young’s Modulus l SpringF = k x è What happens to “k” if cut spring in half? è 1) ½ 2) same3) 2 l k is inversely proportional to length! l Define  Strain =  L / L è Stress = F/A l Now è Stress = Y Strain  F/A = Y  L/L è k = Y A/L from F = k x l Y (Young’s Modules) independent of L 15

What does moving in a circle have to do with moving back & forth in a straight line ?? y x -R R 0  R  x Movie x = R cos  = R cos (  t ) since  =  t 46

(A) (B) (C)

(A) (B) (C)