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Physics 315: Oscillations and Waves Midterm 1. Solutions.
- (a) ω is the characteristic angular oscillation frequency of the system. (b) The characteristic frequency in cycles per second is f = ω/ 2 π. (c) x(t) = x 0 cos(ω t). (d) Now, dE dt =^ 2 ˙x^ x¨^ +^2 ω^
(^2) x x˙, (1) or (^) dE dt =^ 2 ˙x^
x¨ + ω 2 x
However, ¨x + ω 2 x = 0. Hence, dE/dt = 0. E is twice the energy per unit mass of the system.
- We have x¨ + ν x˙ + ω 02 x = 0 , (3) and x(t) = a e−ν^ t/^2 cos(ω 1 t − φ), (4) where ω 1 = (ω 02 − ν 2 /4)^1 /^2 , (a) Now, x(0) = x 0 = a cos φ, (5) and x˙(0) = v 0 = − 2 ν a cos φ + ω 1 a sin φ. (6) The previous two equations imply that a cos φ = x 0 , (7) a sin φ = v^0 +^ ων^1 x^0 /^2. (8) Taking the square root of the sum of the squares of the previous two equations, we obtain a =
[
x 02 + (v^0 +^ ν^ x^0 /2)
2 ω 12
] 1 / 2
Taking the ratio of the two equations, we get tan φ = v^0 + ω^1 ν xx 00 /^2. (10)
(b) Multiplying Eq. (3) by ˙x, and rearranging, we get x˙ x¨ + ω 02 x x˙ = −ν x˙ 2. (11) However, if E = 12 x˙ 2 + 12 ω 02 x 2 (12) then (^) dE dt =^ x˙^ x¨^ +^ ω^ 02 x x˙.^ (13) Hence, (^) dE dt =^ −ν^ x˙^
Now, E is proportional to the energy of the system. Moreover, the right-hand side of the previous equation is always negative. Hence, the previous equation implies that the energy of the system decreases monotonically in time as a consequence of the damping.
- (a) By analogy with the derivation of Eqs. (3.1) and (3.2) of the book, the equations of motion of the two masses are: m ¨x 1 = −k′^ x 1 + k (x 2 − x 1 ), (15) m ¨x 2 = −k (x 2 − x 1 ) + k′^ (−x 2 ). (16) These can be rewritten as x¨ 1 = −(1 + α) ω 02 x 1 + ω 02 x 2 , (17) x¨ 2 = ω 02 x 1 − (1 + α) ω 02 x 2 , (18) where ω 0 = √k/m and α = k′/k. (b) Let us search for a normal mode of the form x 1 (t) = xˆ 1 cos(ω t − φ), (19) x 2 (t) = xˆ 2 cos(ω t − φ). (20) Substitution into the final pair of equations in part (a) yields ( (^) ωˆ 2 − (1 + α) 1 1 ωˆ 2 − (1 + α)
) ( (^) xˆ 1 xˆ 2
where ˆω = ω/ω 0. Setting the determinant of the matrix to zero, we obtain ( ωˆ 2 − [1 + α]
ωˆ 2 − α
ωˆ 2 − [2 + α]
The positive roots of this polynomial are ˆω = √α and ˆω = √ 2 + α. Hence, the normal frequencies are ω = √α ω 0 , (23) and ω 1 = √ 2 + α ω 0. (24)
