CAIE FP2 2012 June — Question 3 8 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2012
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSimple Harmonic Motion
TypeTime to travel between positions
DifficultyChallenging +1.2 This is a multi-step SHM problem requiring application of the standard velocity formula v² = ω²(a² - x²) at two positions to find amplitude and period, followed by integration or time formula application to find travel time. While it involves several steps and algebraic manipulation, the techniques are standard for Further Maths SHM questions with no novel insight required—moderately above average difficulty.
Spec4.10f Simple harmonic motion: x'' = -omega^2 x
3 The point \(O\) is on the fixed line \(l\). Points \(A\) and \(B\) on \(l\) are such that \(O A = 6 \mathrm {~m}\) and \(O B = 8 \mathrm {~m}\), with \(O\) between \(A\) and \(B\). A particle \(P\) oscillates on \(l\) in simple harmonic motion with centre \(O\). When \(P\) is at \(A\) its speed is \(4 \mathrm {~m} \mathrm {~s} ^ { - 1 }\), and when \(P\) is at \(B\) its speed is \(3 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Show that the amplitude of the motion is 10 m and find the period of the motion. Find the time taken by \(P\) to travel directly from \(A\) to \(B\), through \(O\).
Question 3:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(4^2 = \omega^2(A^2 - 6^2)\), \(3^2 = \omega^2(A^2 - 8^2)\)M1 Apply \(v^2 = \omega^2(A^2 - x^2)\) at both points
\(7A^2 = 32^2 - 18^2\), \(A = 10\) A.G.M1 A1 Combine to find \(A\)
\(\omega = \frac{1}{2}\), \(T = 2\pi/\omega = 4\pi\) or \(12.6\) [s]B1 Find period \(T\) using \(\omega\)
*EITHER:* \(t_{OA} = \omega^{-1}\sin^{-1}(6/10) = [1.287]\)M1 Use \(x = A\sin\omega t\) from \(O\) to \(A\)
\(t_{OB} = \omega^{-1}\sin^{-1}(8/10) = [1.855]\)M1 Use \(x = A\sin\omega t\) from \(O\) to \(B\)
\(t_{OA} + t_{OB} = 3.14\) or \(\pi\) [s]M1 A1 Combine to find time from \(A\) to \(B\)
*OR:* \(t_A = \omega^{-1}\cos^{-1}(6/10) = [1.855]\)(M1) Use \(x = A\cos\omega t\) to find time to \(A\)
\(t_B = \omega^{-1}\cos^{-1}(-8/10) = [4.996]\)(M1) Use \(x = A\cos\omega t\) to find time to \(B\)
\(t_B - t_A = 3.14\) or \(\pi\) [s](M1 A1) Combine to find time from \(A\) to \(B\)
Total: 8 marks
## Question 3:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $4^2 = \omega^2(A^2 - 6^2)$, $3^2 = \omega^2(A^2 - 8^2)$ | M1 | Apply $v^2 = \omega^2(A^2 - x^2)$ at both points |
| $7A^2 = 32^2 - 18^2$, $A = 10$ **A.G.** | M1 A1 | Combine to find $A$ |
| $\omega = \frac{1}{2}$, $T = 2\pi/\omega = 4\pi$ or $12.6$ [s] | B1 | Find period $T$ using $\omega$ |
| *EITHER:* $t_{OA} = \omega^{-1}\sin^{-1}(6/10) = [1.287]$ | M1 | Use $x = A\sin\omega t$ from $O$ to $A$ |
| $t_{OB} = \omega^{-1}\sin^{-1}(8/10) = [1.855]$ | M1 | Use $x = A\sin\omega t$ from $O$ to $B$ |
| $t_{OA} + t_{OB} = 3.14$ or $\pi$ [s] | M1 A1 | Combine to find time from $A$ to $B$ |
| *OR:* $t_A = \omega^{-1}\cos^{-1}(6/10) = [1.855]$ | (M1) | Use $x = A\cos\omega t$ to find time to $A$ |
| $t_B = \omega^{-1}\cos^{-1}(-8/10) = [4.996]$ | (M1) | Use $x = A\cos\omega t$ to find time to $B$ |
| $t_B - t_A = 3.14$ or $\pi$ [s] | (M1 A1) | Combine to find time from $A$ to $B$ |

**Total: 8 marks**

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3 The point $O$ is on the fixed line $l$. Points $A$ and $B$ on $l$ are such that $O A = 6 \mathrm {~m}$ and $O B = 8 \mathrm {~m}$, with $O$ between $A$ and $B$. A particle $P$ oscillates on $l$ in simple harmonic motion with centre $O$. When $P$ is at $A$ its speed is $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, and when $P$ is at $B$ its speed is $3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Show that the amplitude of the motion is 10 m and find the period of the motion.

Find the time taken by $P$ to travel directly from $A$ to $B$, through $O$.

\hfill \mbox{\textit{CAIE FP2 2012 Q3 [8]}}
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Q1 7 Q2 7 Q3 8 Q4 11 Q5 11 Q6 7 Q7 7 Q8 9 Q9 9 Q10 12 Q11 EITHER Q11 OR