FACULTY OF SCIENCE, AUTUMN 2007
PHYS 124 LEC A1 : Particles and Waves (Instructor: Marc de Montigny)
Walker, Physics, Chapter 13: Oscillations about Equilibrium
- Section 13-1: Periodic Motion
- P. 395 : Period T = time required for one cycle of a periodic motion [SI Unit : seconds/cycle = s]
- P. 395 : Frequency f = 1/T is the number of cycles per unit of time [SI Unit : cycle/second = 1/s = s-1 or Hertz (Hz)]
- P. 396 Table 13-1 (Typical Periods and Frequencies)
- Section 13-2: Simple Harmonic Motion
- P. 396 (bottom) : restoring force F = -kx
- P. 397 Figure 13-1
- P. 398, Eq. 13-2 : x = A cos(2πt/T)
- P. 425, Problem 15
- Section 13-3: Connections Between Uniform Circular Motion and Simple Harmonic Motion
- Simple Harmonic Oscillator (To obtain Eqs. 13-3 and 13-4, define the x-axis as being vertical.)
- P. 400, Figure 13-5 shows how to obtain Eq. 13-4: x = A cos(ω t)
- Note that the amplitude of oscillation, A, is seen as the radius of the circle, and &omega t as the angle (P. 400, Eq. 13-3).
- P. 400, Eq. 13-5 : ω = 2πf [Unit : rad/second, as before]
- P. 401, Figure 13-6 shows how to obtain Eq. 13-6: v = -ωA sin(ω t), where v is in m/s.
- P. 401, Eq. 13-7 : vmax = Aω. This velocity is attained when the mass is at the equilibrium position.
- P. 402, Figure 13-7 shows how to obtain Eq. 13-8: a = -ω² A cos(ω t), where a is in m/s².
- P. 402, Eq. 13-9 : amax = Aω². This is attained when the mass is at its maximum displacements.
- P. 425, Problem 17
- P. 425, Problem 21
- Section 13-4: The Period of a Mass on a Spring
- P. 405, Eq. 13-10 : ω = (k/m)1/2 (follows from F = ma = -kx, together with Eqs. 13-4 and 13-8.)
- P. 426, Problem 30,Figure 13-22
- P. 426, Problem 37
- Section 13-5: Energy Conservation in Oscillatory Motion
- P. 409, Eq. 13-13 : E = K + U = ½mv² + ½kx²
- P. 410, Eq. 13-15 : E = Umax = ½kA²
- P. 410, Figure 13-10
- P. 410, Eq. 13-16 : E = Kmax = ½mA²ω²
- P. 411, Figure 13-11
- P. 426, Problem 42
- P. 426, Problem 46
- Section 13-6: The Pendulum
- P. 414, Figure 13-14
- P. 414, Eq. 13-20 : T = 2π (L/g)½, which is equivalent to ω = (g/L)½. Proof p. 414.
- P. 429, Problem 79, Figure 13-30