Simple Harmonic Motion Things that vibrate § 14.1–14.3

Hooke’s Law Force Law: F = –kx F = force exerted by the spring k = spring constant (characteristic of the particular spring) x = distance spring is displaced from equilibrium

Conditions of Motion Newton’s second law: F = ma Force F depends on position by Hooke’s law: F = –kx –kx = ma tells how motion changes at each position second-order ordinary differential equation –kx = m d2xd2x dt 2

Think Question The net force on a Hooke’s law object is A.zero at the top and bottom B.maximum at the top and bottom C.minimum but not zero at the top and bottom

Think Question The acceleration of a Hooke’s law object is A.zero at the top and bottom B.maximum at the top and bottom C.minimum but not zero at the top and bottom

Think Question The speed of a Hooke’s law object is A.maximum at the equilibrium position B.maximum at the top and bottom C.maximum midway between equilibrium and top or bottom

Poll Question All other things being equal, if the mass on a Hooke’s law oscillator increases, its period A.decreases. B.does not change. C.increases.

Poll Question All other things being equal, if the stiffness k of a Hooke’s law spring increases, its period A.decreases. B.does not change. C.increases.

Uniform Circular Motion Centripetal force F = mv 2 /r inwards Constant magnitude F 0 ; direction depends on position  F0F0 F0F0 F0F0 F0F0 F0F0 F0F0 Force in y-direction is proportional to –y

Uniform Circular Motion Angle changes at a steady rate. Projection on y-axis has Hooke’s law force. So, projection on y-axis must have Hooke’s law motion too! What is the projection of an angle on the y-axis?

Board Work Verify that if x = A cos (  t +  ), where A and  are any real constants –kx = m d2xd2x dt 2  = k/m

Equations of Motion x(t) = A cos (  t +  ), Amplitude A Angular frequency  Initial phase angle  v = dx/dt a = dv/dt = d 2 x/dt 2

Another form x(t)= A cos (  t +  ) = C cos (  t) + S sin (  t) where C = A cos (  ) and S = –A sin (  ) and If C > 0,  = arctan(–S/C) If C < 0,  = arctan(–S/C) +  A = C 2 + S 2

Period and Frequency Period T –time of one cycle (units: s) Frequency f –cycles per unit time (units: cycles/s = Hz) –f = 1/T Angular frequency  – radians per unit time (units: 1/s or rad/s) –  = 2  f = 2  /T –  2 = k/m

Think Question The potential energy of an oscillating mass is greatest A.at its extreme positions. B.at its equilibrium (middle) position. C.between the middle and an extreme position.

Think Question The kinetic energy of an oscillating mass is greatest A.at its extreme positions. B.at its equilibrium (middle) position. C.between the middle and an extreme position.

Energy Potential energy of a stretched spring : PE = kx Conservation of energy: (This of course ignores the sullen reality of energy dispersal by friction and drag. We’ll get to that.) PE + KE = constant = kA where A is the oscillation amplitude.

Initial Conditions Given m, k, x 0, and v 0, how does one find the equations of motion? m and k give . –  2 = k/m x 0, v 0, and  give A. – 1/2 kA 2 = 1/2 kx /2 mv 0 2

Initial Conditions Given m, k, x 0, and v 0, how does one find the equations of motion? x 0 /A and v 0 /(A  ) give . – x 0 = A cos(  ) – v 0 = –A  sin (  ) – tan(  ) = –v 0 /  x 0 – if x 0 > 0,  = arctan(–v 0 /  x 0 ) – if x 0 < 0,  =  + arctan(–v 0 /  x 0 )

Effect of Gravity Less than you might expect: Changes equilibrium position x = 0 Does not change k

Spring + Gravity position force 0 0 spring alone gravity

Spring + Gravity position force 0 0 spring alone gravity 0 spring + gravity

Spring + Gravity position net force 0 0 different equilibrium position same k