Year: 1991
Paper: 2
Question Number: 12
Course: UFM Mechanics
Section: Simple Harmonic Motion
No solution available for this problem.
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
A particle is attached to one end $B$ of a light elastic string of
unstretched length $a$. Initially the other end $A$ is at rest and
the particle hangs at rest at a distance $a+c$ vertically below $A$.
At time $t=0$, the end $A$ is forced to oscillate vertically, its
downwards displacement at time $t$ being $b\sin pt$. Let $x(t)$
be the downwards displacement of the particle at time $t$ from its
initial equilibrium position. Show that, while the string remains
taut, $x(t)$ satisfies
\[
\frac{\mathrm{d}^{2}x}{\mathrm{d}t^{2}}=-n^{2}(x-b\sin pt),
\]
where $n^{2}=g/c$, and that if $0 < p < n$, $x(t)$ is given by
\[
x(t)=\frac{bn}{n^{2}-p^{2}}(n\sin pt-p\sin nt).
\]
Write down a necessary and sufficient condition that the string remains
taut throughout the subsequent motion, and show that it is satisfied
if $pb < (n-p)c.$