Challenging +1.8 This is a challenging Further Maths question requiring: (1) deriving SHM from non-standard forces using binomial approximation, (2) applying energy conservation with non-elementary integrals or SHM formulas, and (3) solving for time using inverse trig functions. The setup is conceptually demanding and requires multiple sophisticated techniques, placing it well above average difficulty but not at the extreme end for FP2.
4 A particle \(P\) of mass \(m\) moves along part of a horizontal straight line \(A B\). The mid-point of \(A B\) is \(O\), and \(A B = 4 a\). At time \(t , A P = 2 a + x\). The particle \(P\) is acted on by two horizontal forces. One force has magnitude \(m g \left( \frac { 2 a + x } { 2 a } \right) ^ { - \frac { 1 } { 2 } }\) and acts towards \(A\); the other force has magnitude \(m g \left( \frac { 2 a - x } { 2 a } \right)\) and acts towards \(B\). Show that, provided \(\frac { x } { a }\) remains small, \(P\) moves in approximate simple harmonic motion with centre \(O\), and state the period of this motion.
At time \(t = 0 , P\) is released from rest at the point where \(x = \frac { 1 } { 20 } a\). Show that the speed of \(P\) when \(x = \frac { 1 } { 40 } a\) is \(\frac { 1 } { 80 } \sqrt { } ( 3 a g )\), and find the value of \(t\) when \(P\) reaches this point for the first time.
4 A particle $P$ of mass $m$ moves along part of a horizontal straight line $A B$. The mid-point of $A B$ is $O$, and $A B = 4 a$. At time $t , A P = 2 a + x$. The particle $P$ is acted on by two horizontal forces. One force has magnitude $m g \left( \frac { 2 a + x } { 2 a } \right) ^ { - \frac { 1 } { 2 } }$ and acts towards $A$; the other force has magnitude $m g \left( \frac { 2 a - x } { 2 a } \right)$ and acts towards $B$. Show that, provided $\frac { x } { a }$ remains small, $P$ moves in approximate simple harmonic motion with centre $O$, and state the period of this motion.
At time $t = 0 , P$ is released from rest at the point where $x = \frac { 1 } { 20 } a$. Show that the speed of $P$ when $x = \frac { 1 } { 40 } a$ is $\frac { 1 } { 80 } \sqrt { } ( 3 a g )$, and find the value of $t$ when $P$ reaches this point for the first time.
\hfill \mbox{\textit{CAIE FP2 2013 Q4 [9]}}