Chapter 3: Oscillations Introduction: Oscillatory motion in 1dimension. A mass m moves in an arbitrary conservative force field, subject to a potential U(x): F(x)  -(dU/dx) Assume: A position of stable equilibrium (x0) exists.  If m is displaced in either direction from x0: The force F will act to restore it to its original position (x0). Near x0, U(x)  a parabola (There is positive curvature near x0).

Schematically: U(x) near x0 looks (in comparison to a true parabola) like (from Arya’s book)

Consider the restoring force F = F(x) Assume F(x) is “well behaved”:  Derivatives of all orders exist. For simplicity, take the equilibrium position x0 = 0  Expand F(x) in a Taylor’s series about x0 = 0 F(x)  F0 +x(dF/dx)0 +(x2/2!)(d2F/dx2)0 + (x3/3!)(d3F/dx3)0 + (x4/4!)(d4F/dx4)0 + … where F0 = F(0) = (value of F at the equilibrium point) and (dnF/dxn)0 = (value of the nth derivative of F at the equilibrium point)

Recall the definition of equilibrium: F0 = F(0) = 0 Otherwise, the particle isn’t in equilibrium at x = x0 = 0 Consider “small” displacements from x = 0:  Neglect all terms higher order in x than the 1st.  F(x)  -kx where k  - (dF/dx)0 Note: The restoring force F(x) is always directed towards the equilibrium position  (dF/dx)0 < 0  The “Force Constant” (or “spring constant”) is positive: k > 0

Assume U(x) is “well behaved”:  Derivatives of all orders exist. If, instead of expanding F(x), we expand the potential U(x) about the equilibrium point, get equivalent results: Assume U(x) is “well behaved”:  Derivatives of all orders exist. For simplicity, take the equilibrium position x0 = 0  Expand U(x) in a Taylor’s series about x0 = 0 U(x)=U0+x(dU/dx)0+(x2/2!)(d2U/dx2)0 + (x3/3!)(d3U/dx3)0 + (x4/4!)(d4U/dx4)0 + … where U0 = U(0) = (value of U at the equilibrium point) and (dnU/dxn)0 = (value of the nth derivative of U at the equilibrium point)

 U(x)  (½)kx2 where k  (d2U/dx2)0= - (dF/dx)0 In the potential energy, we can take U0 = U(0) = 0 with no loss of generality since the zero of energy is arbitrary! Definition of equilibrium: F0 = F(0)  -(dU/dx)0 = 0 Otherwise, the particle isn’t at equilibrium at x = x0 = 0 Consider “small” displacements from x = 0: Neglect all terms higher order in x than than the 2nd.  U(x)  (½)kx2 where k  (d2U/dx2)0= - (dF/dx)0 Note: the curvature of U(x) is always upward (positive): (d2U/dx2)0 > 0  The “Force Constant” (or “spring constant”) is positive: k > 0

All of this boils down to making the approximation: All of this boils down to making the approximation: U(x)  (½)kx2, F(x) = -(dU/dx)  -kx with k  (d2U/dx2)0 That is, we replace the solid curve with the dashed curve. This is clearly OK in the region not too far from x = x0 = 0 U(x)

Hooke’s “Law” Systems with a restoring force which is linear in the displacement: F(x)  -kx, U(x)  (½)kx2 are said to obey Hooke’s “Law” (Hooke’s Relation) Large numbers of systems obey this relation (at least approximately) as long as the displacement is small. Systems which obey this relation are said to undergo “Elastic” deformations. Some of these are Masses + stretched springs, Pendula, (Small) bending of beams, AC electrical circuits, Others …. Obviously this relation is Approximate! Real systems usually have (especially for displacements which aren’t small) Non-linear restoring forces! (Ch. 4)

From Arya’s Book F(x)  -kx - εx3 U(x)  (½)kx2 + (¼)εx4

Systems for which F(x)  -kx, U(x)  (½)kx2 are said to obey Hooke’s “Law”. The prototype system for this is the Mass-Spring system  “Simple Harmonic Oscillator” Discussed in detail in Physics I. Here, we will discuss it again. We will also add damping (frictional forces) The “Damped Harmonic Oscillator” We will also discuss external driving forces (forced oscillations; resonance) The “Driven Harmonic Oscillator”

Simple Harmonic Oscillator Sect. 3.2 Prototype Simple Harmonic Oscillator (SHO): The Mass - Spring system.

Simple Harmonic Oscillator: Stretch the spring a distance A & release it: In the absence of friction, oscillations go on forever. Newton’s 2nd Law eqtn: F = -kx = mx = ma Define: (ω0)2  k/m  x + (ω0)2 x = 0 A standard 2nd order time dependent differential equation! See Appendix C for solution!

x + (ω0)2 x = 0 General Solution: x(t) = A sin(ω0t - δ) Or: x(t) = A cos(ω0t - φ) Equivalent solutions! Sinusoidal with time. δ = φ + π/2 A = Amplitude of motion. δ, φ = phases A, δ, φ determined by the initial conditions For example, take x(t) = A sin(ω0t - δ)  v(t) = x(t) = (dx/dt) = ω0A cos(ω0t - δ) and a(t) = x(t) = (d2x/dt2) = -(ω0)2A sin(ω0t - δ) = -(ω0)2x(t) Using (ω0)2 = k/m, this gives ma = - kx (N’s 2nd Law again!)

Simple Harmonic Oscillator: Hooke’s “Law” for a vertical spring (take + x as down): Static equilibrium: ∑Fx = 0 = mg - kx0 or x0 = (mg/k) Newton’s 2nd Law equation of motion is the same as before, but the equilibrium position is x0 instead of x = 0

Period & Frequency Period of motion  τ0  Time interval for motion to repeat. x(t) = A sin(ω0t - δ)  x(t+τ0)  Clearly τ0 is such that ω0τ0 = 2π Or: τ0 = 2π/ω0 But: (ω0)2  k/m So, the period is: τ0= 2π(m/k)½ The physical significance of ω0: It is the angular frequency of the motion. It is related to the frequency ν0 by: ω0 = 2πν0 & ν0 = 1/τ0 = (2π)-1(k/m)½ Units: [ν0] = Hertz (Hz), [ω0] = Rad/s The period τ0 of a SHO is independent of amplitude A!

Suppose, initial conditions: at t = 0, x = x0, v = v0 x(t) = A sin(ω0t - δ) Amplitude A, & phase δ given by initial conditions, as follows: v(t) = (dx/dt) = ω0A cos(ω0t - δ) Suppose, initial conditions: at t = 0, x = x0, v = v0  x0= - A sin(δ); v0= ω0A cos(δ) Combining, these gives: cot(δ) = v0/(ω0x0) and A = [(x0)2 + (v0)2/(ω0)2]½ Special cases: 1. At t = 0, x = x0, v = 0  A= x0, δ = -π/2  x(t) = x0cos(ω0t), v(t) = ω0x0 sin(ω0t) 2. At t = 0, x = 0, v = v0  A = v0/ω0, δ = 0  x(t) = (v0/ω0)sin(ω0t), v(t) = v0cos(ω0t)

Plot of x(t) (δ/ω0)

Plots of x(t), v(t), a(t) for δ = ½π

Energy Kinetic Energy: T = (½)mv2 = (½)mx2 v(t)= ω0Acos(ω0t - δ)  T = T(t) = (½)m(ω0A)2cos2(ω0t - δ) Potential Energy: U(x) = (½)kx2 x(t) = A sin(ω0t - δ)  U = U(t) = ½kA2sin2(ω0t - δ) Total Energy (note that (ω0)2  (k/m)) : E = T + U = (½)m(ω0A)2cos2(ω0t - δ) + (½)kA2sin2(ω0t - δ)  E = (½)kA2 [cos2(ω0t - δ) + sin2(ω0t - δ)] = Or: E = (½)kA2 a constant, independent of time!

But E = constant  (½)kA2 = (½)m(v0)2 or v0 = ± (k/m)½A = ± ω0A Total Energy: E =T + U = (½)mv2 +(½)kx2 = constant x = A, v = 0, T = 0 E = U = (½)kA2  All potential energy! x = 0, v = v0, U = 0 E = T = (½)m(v0)2  All kinetic energy! But E = constant  (½)kA2 = (½)m(v0)2 or v0 = ± (k/m)½A = ± ω0A

Partially kinetic energy Total Energy: E =T + U = (½)mv2 + (½)kx2 = constant x = -A, v = 0, T = 0 E = U = (½)kA2  All potential energy! In between: -A  x  A E = T + U = (½)mv2 + (½)kx2  Partially kinetic energy Partially potential energy But E = constant  (½)kA2 = (½)m(v0)2 = (½)mv2 + (½)kx2 So (for example): v = ± (k/m)½[A2 –x2]½ = ± ω0[A2 –x2]½

Example 3.1: Physical Pendulum Find the frequency & the period of oscillation of a uniform solid sphere, mass m, radius R about a point on its surface. (Assume small oscillations). I  moment of inertia about pivot point Physics I results: Moment of inertia about center = (2mR2/5). Parallel axis theorem: I = (2mR2/5) + mR2 = (7mR2/5) Clearly, F = mg. Torque about pivot point N = R × F. Or |N| = Rmg sinθ N’s 2nd Law eqtn of motion (rotations): I θ = -|N| = - Rmg sinθ Small oscillations  sinθ  θ  I θ  - Rmg θ

Equation of motion for sphere (small oscillations): I θ + Rmg θ = 0 Or: (7mR2/5) θ + Rmg θ = 0 Simplifying: θ + (5g/7R)θ = 0 (1) NOTE: The simple harmonic oscillator eqtn of motion is: x + (ω0)2 x = 0 (2) (1) & (2) are identical mathematically!  We identify the angular frequency of the sphere by direct comparison of (1) & (2). That is: (ω0)2 = (5g/7R) or ω0 = (5g/7R)½ Similarly the period of oscillation is: τ = (2π/ω0) = 2π(7R/5g)½