Simple Nature - Light and Matter

This looks like a cosine function, so let’s see if a x = A cos (ωt+δ)is a solution to the conservation of energy equation — it’s not uncommonto try to “reverse-engineer” the cryptic results of a numericalcalculation like this. The symbol ω = 2π/T (Greek omega),called angular frequency, is a standard symbol for the number ofradians per second of oscillation. Except for the factor of 2π, it isidentical to the ordinary frequency f = 1/T , which has units of s −1or Hz (Hertz). The phase angle δ is to allow for the possibility thatt = 0 doesn’t coincide with the beginning of the motion. The energyisE = K + U= 1 2 mv2 + 1 2 kx2= 1 ( ) dx 22 m + 1 dt 2 kx2= 1 2 m [−Aω sin (ωt + δ)]2 + 1 k [A cos (ωt + δ)]22= 1 [ 2 A2 mω 2 sin 2 (ωt + δ) + k cos 2 (ωt + δ) ]According to conservation of energy, this has to be a constant. Usingthe identity sin 2 + cos 2 = 1, we can see that it will be a constant ifwe have mω 2 = k, or ω = √ k/m, i.e., T = 2π √ m/k. Note that theperiod is independent of amplitude.d / Example 23. The rod pivotson the hinge at the bottom.A spring and a lever example 23⊲ What is the period of small oscillations of the system shown inthe figure? Neglect the mass of the lever and the spring. Assumethat the spring is so stiff that gravity is not an important effect.The spring is relaxed when the lever is vertical.⊲ This is a little tricky, because the spring constant k, althoughit is relevant, is not the k we should be putting into the equationT = 2π √ m/k. The k that goes in there has to be the secondderivative of U with respect to the position, x, of the mass that’smoving. The energy U stored in the spring depends on how farthe tip of the lever is from the center. This distance equals (L/b)x,116 Chapter 2 Conservation of Energy