CAIE FP2 2013 November — Question 2 8 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2013
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSimple Harmonic Motion
TypeTime to travel between positions
DifficultyChallenging +1.2 This is a standard SHM problem requiring application of the velocity formula v² = ω²(a² - x²) to find amplitude and speed, followed by integration to find time between positions. While it involves multiple steps and careful algebraic manipulation, the techniques are routine for Further Maths students and follow predictable patterns without requiring novel insight.
Spec6.05f Vertical circle: motion including free fall

2 The point \(O\) is on the fixed line \(l\). The point \(A\) on \(l\) is such that \(O A = 3 \mathrm {~m}\). A particle \(P\) oscillates on \(l\) in simple harmonic motion with centre \(O\) and period \(\pi\) seconds. When \(P\) is at \(A\) its speed is \(12 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Find the speed of \(P\) when it is at the point \(B\) on \(l\), where \(O B = 6 \mathrm {~m}\) and \(B\) is on the same side of \(O\) as \(A\). Find, correct to 2 decimal places, the time, in seconds, taken for \(P\) to travel directly from \(A\) to \(B\).

Question 2:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\omega = 2\)B1 Use \(T = \frac{2\pi}{\omega}\) to find \(\omega\)
\(12^2 = \omega^2(A^2 - 3^2)\), \(A^2 = 45\)M1 A1 Use \(v^2 = \omega^2(A^2 - x^2)\) to find \(A^2\)
\(v_B^2 = 2^2(45 - 6^2)\), \(v_B = 6\) ms\(^{-1}\)B1\(\checkmark\) Use \(v^2 = \omega^2(A^2 - x^2)\) to find \(v\) at \(x=6\)
\(\frac{1}{2}\sin^{-1}(3/\sqrt{45})\) or \(\frac{1}{2}\cos^{-1}(3/\sqrt{45})\) Find time at \(A\)
\(\frac{1}{2}\sin^{-1}(6/\sqrt{45})\) or \(\frac{1}{2}\cos^{-1}(6/\sqrt{45})\)M1 A1 or at \(B\)
\(\frac{1}{2}\sin^{-1}(6/\sqrt{45}) - \frac{1}{2}\sin^{-1}(3/\sqrt{45})\) Combine correctly to find time from \(A\) to \(B\)
\(= 0.554 - 0.232 = 0.32\) sM1, A1\(\checkmark\) Evaluate to 2 d.p.; \([\checkmark\) on \(\omega]\)
Total: 8 marks
## Question 2:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\omega = 2$ | B1 | Use $T = \frac{2\pi}{\omega}$ to find $\omega$ |
| $12^2 = \omega^2(A^2 - 3^2)$, $A^2 = 45$ | M1 A1 | Use $v^2 = \omega^2(A^2 - x^2)$ to find $A^2$ |
| $v_B^2 = 2^2(45 - 6^2)$, $v_B = 6$ ms$^{-1}$ | B1$\checkmark$ | Use $v^2 = \omega^2(A^2 - x^2)$ to find $v$ at $x=6$ |
| $\frac{1}{2}\sin^{-1}(3/\sqrt{45})$ or $\frac{1}{2}\cos^{-1}(3/\sqrt{45})$ | | Find time at $A$ |
| $\frac{1}{2}\sin^{-1}(6/\sqrt{45})$ or $\frac{1}{2}\cos^{-1}(6/\sqrt{45})$ | M1 A1 | or at $B$ |
| $\frac{1}{2}\sin^{-1}(6/\sqrt{45}) - \frac{1}{2}\sin^{-1}(3/\sqrt{45})$ | | Combine correctly to find time from $A$ to $B$ |
| $= 0.554 - 0.232 = 0.32$ s | M1, A1$\checkmark$ | Evaluate to 2 d.p.; $[\checkmark$ on $\omega]$ |
| **Total: 8 marks** | | |

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2 The point $O$ is on the fixed line $l$. The point $A$ on $l$ is such that $O A = 3 \mathrm {~m}$. A particle $P$ oscillates on $l$ in simple harmonic motion with centre $O$ and period $\pi$ seconds. When $P$ is at $A$ its speed is $12 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Find the speed of $P$ when it is at the point $B$ on $l$, where $O B = 6 \mathrm {~m}$ and $B$ is on the same side of $O$ as $A$.

Find, correct to 2 decimal places, the time, in seconds, taken for $P$ to travel directly from $A$ to $B$.

\hfill \mbox{\textit{CAIE FP2 2013 Q2 [8]}}