CAIE FP2 2013 June — Question 4 9 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2013
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSimple Harmonic Motion
TypeSmall oscillations with non-standard force laws
DifficultyChallenging +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.
Spec4.10f Simple harmonic motion: x'' = -omega^2 x6.06a Variable force: dv/dt or v*dv/dx methods

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.

Question 4:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(m\,d^2x/dt^2 = mg(2a-x)/2a - mg\{(2a+x)/2a\}^{-\frac{1}{2}}\)M1 A1 Find eqn. of motion at general point (wrong sign loses this A1 only)
\(d^2x/dt^2 = g(1-x/2a) - g(1-x/4a) = -gx/4a\)M1 A1 Neglect \((x/a)^2\) and higher powers
\(T = 2\pi\sqrt{4a/g}\) or \(4\pi\sqrt{a/g}\)B1 Find period from \(T = 2\pi/\omega\)
\(v^2 = (g/4a)\{(a/20)^2 - (a/40)^2\}\)M1 Find \(v\) from \(v^2 = \omega^2(A^2 - x^2)\)
\(v = (1/80)\sqrt{3ag}\) A.G.A1
\(t = \sqrt{4a/g}\cos^{-1}\frac{1}{2} = \frac{2}{3}\pi\sqrt{a/g}\)M1 A1 Find \(t\) from \(x = a\cos\omega t\) (A.E.F.)
Total: 9 marks
## Question 4:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $m\,d^2x/dt^2 = mg(2a-x)/2a - mg\{(2a+x)/2a\}^{-\frac{1}{2}}$ | M1 A1 | Find eqn. of motion at general point (wrong sign loses this A1 only) |
| $d^2x/dt^2 = g(1-x/2a) - g(1-x/4a) = -gx/4a$ | M1 A1 | Neglect $(x/a)^2$ and higher powers |
| $T = 2\pi\sqrt{4a/g}$ or $4\pi\sqrt{a/g}$ | B1 | Find period from $T = 2\pi/\omega$ |
| $v^2 = (g/4a)\{(a/20)^2 - (a/40)^2\}$ | M1 | Find $v$ from $v^2 = \omega^2(A^2 - x^2)$ |
| $v = (1/80)\sqrt{3ag}$ **A.G.** | A1 | |
| $t = \sqrt{4a/g}\cos^{-1}\frac{1}{2} = \frac{2}{3}\pi\sqrt{a/g}$ | M1 A1 | Find $t$ from $x = a\cos\omega t$ (A.E.F.) |

**Total: 9 marks**

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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]}}