 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2013/paper-2/7/ii/solution

The displacement directions specified at release give

x (0) = (- x_{0}, x_{0})^{T}

. Let

Δ r = r_{-} - r_{+} = 2 D / m_{2} > 0

x (0) = c_{-} v_{-} + c_{+} v_{+}

. Solving the two coordinate equations gives

c_{-} = \frac{x _{0} ( 1 + r _{+} )}{Δ r}, c_{+} = - \frac{x _{0} ( 1 + r _{-} )}{Δ r} .

Zero initial velocities remove the sine terms, so

x (t) = \frac{x _{0}}{Δ r} [(1 + r_{+}) v_{-} cos (ω_{-} t) - (1 + r_{-}) v_{+} cos (ω_{+} t)] .

In components,

x_{1} (t) x_{2} (t) = \frac{x _{0}}{Δ r} [(1 + r_{+}) cos (ω_{-} t) - (1 + r_{-}) cos (ω_{+} t)], = \frac{x _{0}}{Δ r} [r_{-} (1 + r_{+}) cos (ω_{-} t) - r_{+} (1 + r_{-}) cos (ω_{+} t)] .

Substitution at zero gives

x_{1} = - x_{0}

x_{2} = x_{0}

, and both derivatives vanish. For unequal masses both normal modes are generally present. When

m_{1} = m_{2} = m

r_{-} = 1

r_{+} = - 1

, so only the higher mode survives:

x_{1} = - x_{0} cos (3 k / m t)

and

x_{2} = x_{0} cos (3 k / m t)

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