Simple Harmonic Motion mass on a spring F = m a Newton’s Law F = – k x spring force m a = – k x a = – (k / m) x acceleration is proportional to x but in opposite direction 

d2x / dt2 = – (k / m) x 2 = k / m Differential Equation x = A cos  t try a solution: dx / dt = – A  sin  t d2x / d2t = – A 2 cos  t = – 2 x this is a solution provided 2 = k / m

A is the amplitude T is the period Plot of Solution • how far I pull the mass down at the start (t = 0) • the maximum displacement T is the period • the time after which the motion repeats

 = 1 / T   = 2  / T = 2   is the frequency • how many time per second the motion repeats cos  t = cos  (t + T) = cos ( t +  T)   T = 2   = 2  / T = 2    is the angular frequency

Initial Conditions x = A cos( t + ) general solution   = k / m  is fixed by k and m  A and  are determined by the initial conditions (how we start the oscillation)

x = A cos ( t + ) xmax = A v = –  A sin ( t + ) vmax =  A occurs when x = 0 a = – 2 A cos ( t + ) amax = 2 A occurs when x = xmax

= B cos  t + C sin  t x0 = B  v0 =  C Another form of the general solution x = A cos( t + ) = A cos  t cos = A cos  t cos – A sin  t sin B = A cos  C = – A sin  = B cos  t + C sin  t B and C are set by initial conditions at t = 0 x0 = B v0 =  C  B is the initial displacement C is the initial velocity / 

d2 / dt2 = – (g / l )  Simple Pendulum l  sin     F = m a = m l  l  F = – m g sin  m l  = – m g sin  m g sin  m g  = – (g / l ) sin  d2 / dt2 = – (g / l ) sin  not equation of SHO but let  be small  sin     d2 / dt2 = – (g / l )  SHO equation

d2x / dt2 = – (k / m) x x = A cos  t  = k / m pendulum motion is approximately SHO the approximation is better for smaller angles Compare d2x / dt2 = – (k / m) x x = A cos  t solution  = k / m with d2 / dt2 = – (g / l )   = A cos  t solution  = g / l

Simple Harmonic Oscillator? What is the Energy of a Simple Harmonic Oscillator? K = ½ m v2 U = ½ k x2 Substitute SHO solution for x, v

= k A2 sin2 t K = m v2 = m 2A2 sin2 t U = k x2 = k A2 cos2 t Energy of Simple Harmonic Oscillation x = A cos  t v = – A sin t K = m v2 = m 2A2 sin2 t = k A2 sin2 t U = k x2 = k A2 cos2 t 1 2 – both K and U oscillate with time

K = k A2 sin2 t 1 2 – 1 2 – U = k A2 cos2 t

Total Energy E = K + U = k A2 = k A2 sin2 t + k A2 cos2 t 1 2 – 1 2 – E = K + U 1 2 – a constant! TOTAL Energy does NOT change with time = k A2 K cannot be negative turning point turning point K

How to Calculate the Force from the Potential Energy U = –  F • dr = –  (Fx dx + Fy dy + Fz dz) dU = – Fx dx – Fy dy – Fz dz Fx = – dU / dx Fy = – dU / dy Fz = – dU / dz the force is minus the slope of the potential F > 0 F < 0 F = 0

Why do we care about SHO? Because any potential energy curve looks (spring and pendulum are fun, but not really fundamental physics…) Because any potential energy curve looks like a simple harmonic oscillator near its minimum. U x Close to min, it looks like SHO Zoom in

Energy Diagram of Diatomic Molecule e.g. O2 let r be atomic separation r r0 equilibrium separation U dU / dr < 0 repulsive U  0 as r  0 r r0 dU / dr > 0 attractive equilibrium point

 U r r0 near equilibrium point looks like SHO approx. parabolic potential atoms displaced from equilibrium behave like simple harmonic oscillator as long as displacement from equilibrium point is not too large for small excitations we can model a diatomic molecule as two point masses connected by a spring 

Complex Numbers introduce i =  – 1 real numbers natural numbers whole numbers integers rational numbers irrational numbers real numbers all algebraic equations cannot be solved using the real numbers e.g. x2 = – 1 introduce i =  – 1

i times a real number is an imaginary number i b e.g. i 5.3 a complex number a + i b is the sum of a real number and an imaginary number e.g. 6.9 + i 5.3 real part imaginary part

One of the Most Remarkable Theorems in all of Mathematics our number system is now complete any algebraic equation has a solution that is a complex number

a + i b Complex Plane = A cos  + i A sin  tan  = b / a any complex number can be represented by a point in an abstract 2 – dimensional space a + i b imaginary axis = A cos  + i A sin   real axis point labeled by A and  tan  = b / a A =  a2 + b2

ei  = cos  + i sin  e is the base of the natural logarithm ln The Euler Relation One of the Most Remarkable Relations in all of Mathematics ei  = cos  + i sin  e is the base of the natural logarithm ln e = 2.7184 • • • e  1 +  for small  from this relation and the properties of exponents we get all of trigonometry

d(A et) / dt =  A et d2(A et) / dt2 = 2 A et Tool for Solving Differential Equations taking derivatives of an exponential function is easy d(A et) / dt =  A et d2(A et) / dt2 = 2 A et its as simple as multiplication

x = Aei t x = A cos t Pretend Solution is a Complex Number rather then then we can substitute it into the differential equation and easily take the derivatives of course we have to remember that the actual physical solution is the real part of our assumed solution Re(Aei t)

d2x / dt2 + (k / m) x = 0 x = Aei t ( – 2 + (k / m)) Aei t = 0 try as a solution: x = Aei t ( – 2 + (k / m)) Aei t = 0 – 2 + (k / m) = 0 2 = (k / m) is the natural frequency 2 = (k / m)

 r = R cos  t A tunnel through the Earth F = – G m M / r2 = – (G m M / R3) r • m a = – (G m M / R3) r d2r / dt2 = – (G M / R3) r SHO equation with  = GM / R3  r = R cos  t

= 84.3 min Same as the time it takes a satellite T = 2  /  = 2   R3 / G M = 2   R/ g = 6.28  6.37 x 106 m / 9.8 m / s2 = 5.06 x 103 s = 84.3 min Same as the time it takes a satellite to orbit the Earth right at the surface

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