CAIE FP2 2011 November — Question 2 7 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2011
SessionNovember
Marks7
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Mark schemeDownload PDF ↗
TopicSimple Harmonic Motion
TypeFind amplitude of SHM
DifficultyStandard +0.3 This is a standard SHM problem requiring application of the velocity formula v² = ω²(a² - x²) at two positions to form simultaneous equations and solve for amplitude. The second part involves using v_max = ωa. While it requires algebraic manipulation and understanding of SHM relationships, it follows a well-established method with no novel insight needed, making it slightly easier than average.
Spec4.10f Simple harmonic motion: x'' = -omega^2 x
2 A particle \(P\) is moving in simple harmonic motion with centre \(O\). When \(P\) is 5 m from \(O\) its speed is \(V \mathrm {~m} \mathrm {~s} ^ { - 1 }\), and when it is 9 m from \(O\) its speed is \(\frac { 3 } { 5 } V \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Show that the amplitude of the motion is \(\frac { 15 } { 2 } \sqrt { } 2 \mathrm {~m}\). Given that the greatest speed of \(P\) is \(3 \sqrt { } 2 \mathrm {~m} \mathrm {~s} ^ { - 1 }\), find \(V\).
Question 2:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Apply \(v^2 = \omega^2(A^2 - x^2)\) at first point: \(V^2 = \omega^2(A^2 - 5^2)\)B1
Apply \(v^2 = \omega^2(A^2 - x^2)\) at second point: \((9/25)V^2 = \omega^2(A^2 - 9^2)\)B1
Combine: \(25(A^2 - 81) = 9(A^2 - 25)\)
\(16A^2 = 25 \times 72\), \(A = 15\sqrt{2}/2\)M1 A1 A.G.
Find \(\omega\): \(\omega = (3\sqrt{2})/(15\sqrt{2}/2) = 2/5\)M1
Find \(V\): \(V^2 = (4/25)(225/2 - 25) = 14\)
\(V = \sqrt{14}\) or \(3.74\)M1 A1
Total: 7 marks
## Question 2:

| Answer/Working | Marks | Guidance |
|---|---|---|
| Apply $v^2 = \omega^2(A^2 - x^2)$ at first point: $V^2 = \omega^2(A^2 - 5^2)$ | B1 | |
| Apply $v^2 = \omega^2(A^2 - x^2)$ at second point: $(9/25)V^2 = \omega^2(A^2 - 9^2)$ | B1 | |
| Combine: $25(A^2 - 81) = 9(A^2 - 25)$ | | |
| $16A^2 = 25 \times 72$, $A = 15\sqrt{2}/2$ | M1 A1 | A.G. |
| Find $\omega$: $\omega = (3\sqrt{2})/(15\sqrt{2}/2) = 2/5$ | M1 | |
| Find $V$: $V^2 = (4/25)(225/2 - 25) = 14$ | | |
| $V = \sqrt{14}$ or $3.74$ | M1 A1 | |

**Total: 7 marks**

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2 A particle $P$ is moving in simple harmonic motion with centre $O$. When $P$ is 5 m from $O$ its speed is $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$, and when it is 9 m from $O$ its speed is $\frac { 3 } { 5 } V \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Show that the amplitude of the motion is $\frac { 15 } { 2 } \sqrt { } 2 \mathrm {~m}$.

Given that the greatest speed of $P$ is $3 \sqrt { } 2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, find $V$.

\hfill \mbox{\textit{CAIE FP2 2011 Q2 [7]}}
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