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Simple Harmonic Motion

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. d 2 x. –kx = m. dt 2. Conditions of Motion.

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Simple Harmonic Motion

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  1. Simple Harmonic Motion Things that vibrate § 14.1–14.3

  2. 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

  3. d2x –kx = m dt2 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-orderordinary differential equation

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

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

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

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

  8. Poll Question All other things being equal, if the stiffnessk of a Hooke’s law spring increases, its period decreases. does not change. increases.

  9. F0 F0 F0 F0 F0 F0 Uniform Circular Motion • Centripetal force F = mv2/r inwards • Constant magnitude F0; direction depends on position q • Force in y-direction is proportional to –y

  10. 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?

  11. w =   k/m d2x –kx = m dt2 Board Work Verify that • if x = A cos (wt + f), • where • A and f are any real constants

  12. Equations of Motion x(t) = A cos (wt + f), • Amplitude A • Angular frequency w • Initial phase angle f • v = dx/dt • a = dv/dt = d2x/dt2

  13. Another form x(t) = A cos (wt + f) = C cos (wt) + S sin (wt) where C = A cos (f) and S = –A sin (f) and A= C2 + S2 If C > 0, f = arctan(–S/C) If C < 0, f = arctan(–S/C) + p

  14. 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 frequencyw • radians per unit time (units: 1/s or rad/s) • w = 2pf = 2p/T • w2 =  k/m

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

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

  17. 1 1 kx2 PE= 2 2 kA2 PE+ KE = constant = Energy • Potential energy of a stretched spring : • Conservation of energy: where A is the oscillation amplitude. (This of course ignores the sullen reality of energy dispersal by friction and drag. We’ll get to that.)

  18. Initial Conditions Given m, k, x0, and v0, how does one find the equations of motion? • m and k give w. • w2 = k/m • x0, v0, and w give A. • 1/2 kA2 = 1/2 kx02 + 1/2 mv02

  19. Initial Conditions Given m, k, x0, and v0, how does one find the equations of motion? • x0/A and v0/(Aw) give f. • x0 = A cos(f) • v0 = –Aw sin (f) • tan(f) = –v0/wx0 • if x0 > 0, f = arctan(–v0/wx0) • if x0 < 0, f = p + arctan(–v0/wx0)

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

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

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

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

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