1996 Paper 2 Q11

Year: 1996
Paper: 2
Question Number: 11

Course: UFM Mechanics
Section: Simple Harmonic Motion

Difficulty: 1600.0 Banger: 1484.0

Problem

A particle hangs in equilibrium from the ceiling of a stationary lift, to which it is attached by an elastic string of natural length \(l\) extended to a length \(l+a\). The lift now descends with constant acceleration \(f\) such that \(0 < f < g/2\). Show that the extension \(y\) of the string from its equilibrium length satisfies the differential equation $$ {{\rm d}^2 y \over {\rm d} t^2} +{g \over a}\, y = g-f. $$ Hence show that the string never becomes slack and the amplitude of the oscillation of the particle is \(af/g\). After a time \(T\) the lift stops accelerating and moves with constant velocity. Show that the string never becomes slack and the amplitude of the oscillation is now \[\frac{2af}{g}|\sin {\textstyle \frac{1}{2}}\omega T|,\] where \(\omega^{2}=g/a\).

No solution available for this problem.

Rating Information

Difficulty Rating: 1600.0

Difficulty Comparisons: 0

Banger Rating: 1484.0

Banger Comparisons: 1

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Problem source
A particle hangs in equilibrium from the ceiling
of a stationary lift, to which it is attached by an elastic
string of natural length $l$ extended to a length $l+a$.
The lift now descends with constant acceleration $f$
such that $0 <  f < g/2$. Show that the extension $y$ of the
string from its equilibrium length satisfies the 
differential equation 
$$
{{\rm  d}^2 y \over {\rm d} t^2}  +{g \over a}\, y = g-f.
$$
Hence show that 
the string never becomes slack and the
amplitude of the oscillation of the particle is $af/g$.
After a time $T$ the lift stops accelerating and moves with constant
velocity. Show that the string never becomes slack
and the amplitude of the oscillation is now
\[\frac{2af}{g}|\sin {\textstyle \frac{1}{2}}\omega T|,\]
where $\omega^{2}=g/a$.