 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2017/paper-1/3/b/solution

The pulse height is the constant

t_{0}

, not the time variable. In the undamped case the convolution therefore reduces to

x (t) = \frac{t _{0}}{2} \int_{0}^{m i n (t, t_{0})} sin [2 (t - τ)] d τ .

Integration gives the continuous displacement and velocity

x (t) = ⎩ ⎨ ⎧ \frac{t _{0}}{4} (1 - cos 2 t), \frac{t _{0}}{4} [cos 2 (t - t_{0}) - cos 2 t], 0 \leq t < t_{0}, t \geq t_{0} .

The post-pulse expression is equivalently

x (t) = (t_{0} /2) sin t_{0} sin (2 t - t_{0})

. Since

t_{0} > 0

, it vanishes for every subsequent time precisely when

sin t_{0} = 0

. Thus

t_{0} = nπ, n = 1, 2, 3, \dots .

These are integer multiples of the oscillator's natural period

π

. At such switch-off times both

x (t_{0})

and

x^{'} (t_{0})

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