1991 Paper 2 Q12

Year: 1991
Paper: 2
Question Number: 12

Course: UFM Mechanics
Section: Simple Harmonic Motion

Difficulty: 1600.0 Banger: 1500.0

Problem

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 \(pb < (n-p)c.\)

No solution available for this problem.

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Difficulty Rating: 1600.0

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Banger Rating: 1500.0

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Problem source
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 $pb < (n-p)c.$