2004 Paper 3 Q10

Year: 2004
Paper: 3
Question Number: 10

Course: UFM Mechanics
Section: Simple Harmonic Motion

Difficulty: 1700.0 Banger: 1484.0
SHM elastic string Hooke's law modulus of elasticity piecewise motion period vertical oscillation two strings

Problem

A particle \(P\) of mass \(m\) is attached to points \(A\) and \(B\), where \(A\) is a distance \(9a\) vertically above \(B\), by elastic strings, each of which has modulus of elasticity \(6mg\). The string \(AP\) has natural length \(6a\) and the string \(BP\) has natural length \(2a\). Let \(x\) be the distance \(AP\). The system is released from rest with \(P\) on the vertical line \(AB\) and \(x = 6a\). Show that the acceleration \(\ddot{x}\) of \(P\) is \(\ds{4g \over a}(7a - x)\) for \(6a < x < 7a\) and \(\ds{g \over a}(7a - x)\) for \(7a < x < 9a\,\). Find the time taken for the particle to reach \(B\).

No solution available for this problem.

Rating Information

Difficulty Rating: 1700.0

Difficulty Comparisons: 0

Banger Rating: 1484.0

Banger Comparisons: 1

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Problem source
A particle $P$ of mass $m$ is attached to points $A$ and $B$, where $A$ is a distance $9a$ 
vertically above $B$, by elastic strings, 
each of which has modulus of elasticity $6mg$. 
The string $AP$ has natural length $6a$ and the string 
$BP$ has natural length $2a$. Let $x$ be the distance $AP$.
The system is released from rest with $P$ 
on the vertical line $AB$ and $x = 6a$. 
Show that the acceleration $\ddot{x}$ of $P$ is 
$\ds{4g \over a}(7a - x)$ for $6a < x < 7a$ 
and 
$\ds{g \over a}(7a - x)$ for $7a < x < 9a\,$.
Find the time taken for the particle to reach $B$.
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