CAIE FP2 2015 June — Question 3 9 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2015
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSimple Harmonic Motion
TypeFind period from given information
DifficultyChallenging +1.2 This is a multi-part SHM question requiring understanding of the relationship between position, velocity, and amplitude, plus solving equations involving trigonometric functions. While it involves several steps and careful setup of SHM equations, the techniques are standard for Further Maths students who have studied SHM systematically. The time-based constraint and final amplitude calculation add moderate complexity but don't require novel insight beyond applying v² = ω²(a² - x²) and standard SHM formulas.
Spec1.05g Exact trigonometric values: for standard angles4.10f Simple harmonic motion: x'' = -omega^2 x
3 A particle moves on a straight line \(A O B\) in simple harmonic motion, where \(A B = 2 a \mathrm {~m}\). The centre of the motion is \(O\) and the particle is instantaneously at rest at \(A\) and \(B\). The point \(M\) is the mid-point of \(O B\). The particle passes through \(M\) moving towards \(O\) and next achieves its maximum speed one second later. Find the period of the motion. Find the distance of the particle from \(O\) when its speed is equal to one half of its maximum speed. At an instant 2.5 seconds after the particle passes through \(M\) moving towards \(O\), the distance of the particle from \(O\) is \(\sqrt { } 2 \mathrm {~m}\). Find, in metres, the amplitude of the motion.
Question 3:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Position where max. speed next attained: Centre OM1 State or imply
\(1 = \omega^{-1}\sin^{-1}\frac{1}{2}\), \(\omega = \pi/6\)M1 A1 EITHER: Find \(\omega\) from SHM eqn. \(x = a\sin\omega t\)
\(T = 2\pi/(\pi/6) = 12\) [s]B1\(\checkmark\) Find period \(T\) from \(T = 2\pi/\omega\)
\(1 = \omega^{-1}\cos^{-1}0 - \omega^{-1}\cos^{-1}\frac{1}{2}\) OR: \(x = a\cos\omega t\)
\(\omega = \pi/2 - \pi/3 = \pi/6\)M1 A1
\(T = 2\pi/(\pi/6) = 12\) [s]B1\(\checkmark\)
\(1 = T/4 - (T/2\pi)\cos^{-1}\frac{1}{2}\)M1 A1 OR: Find \(\omega\) from SHM eqn.
\(= T(1/4 - 1/6)\), \(T = 12\) [s]A1
\(\omega^2(a^2 - x^2) = \frac{1}{4}v_{\max}^2\)M1 Find distance \(x\) from O when speed is half max
\(v_{\max} = a\omega\) [= \(\pi a/6\)]M1 Use maximum speed
\(x^2 = \frac{3}{4}a^2\) note \(x\) is independent of \(\omega\)
\(x = \sqrt{3}a/2\) or \(0.866a\)A1
\(2.5 = \omega^{-1}\sin^{-1}(\sqrt{2}/a) - \omega^{-1}\sin^{-1}(-\frac{1}{2})\) EITHER: Find \(a\) from \(x = a\sin\omega t\)
or \(1.5 = \omega^{-1}\sin^{-1}(\sqrt{2}/a)\)
\(\sin^{-1}(\sqrt{2}/a) = \pi/4\), \(a = 2\) [m]M1 A1
\(2.5 = \omega^{-1}\cos^{-1}(-\sqrt{2}/a) - \omega^{-1}\cos^{-1}\frac{1}{2}\) OR: Find \(a\) from \(x = \cos\omega t\)
or \(1.5 = \omega^{-1}\cos^{-1}(-\sqrt{2}/a) - \omega^{-1}\cos^{-1}0\)
\(\cos^{-1}(-\sqrt{2}/a) = 3\pi/4\), \(a = 2\) [m]M1 A1
Part marks: 4, 3, 2Total: 9 
## Question 3:

| Answer/Working | Marks | Guidance |
|---|---|---|
| Position where max. speed next attained: Centre O | M1 | State or imply |
| $1 = \omega^{-1}\sin^{-1}\frac{1}{2}$, $\omega = \pi/6$ | M1 A1 | EITHER: Find $\omega$ from SHM eqn. $x = a\sin\omega t$ |
| $T = 2\pi/(\pi/6) = 12$ [s] | B1$\checkmark$ | Find period $T$ from $T = 2\pi/\omega$ |
| $1 = \omega^{-1}\cos^{-1}0 - \omega^{-1}\cos^{-1}\frac{1}{2}$ | | OR: $x = a\cos\omega t$ |
| $\omega = \pi/2 - \pi/3 = \pi/6$ | M1 A1 | |
| $T = 2\pi/(\pi/6) = 12$ [s] | B1$\checkmark$ | |
| $1 = T/4 - (T/2\pi)\cos^{-1}\frac{1}{2}$ | M1 A1 | OR: Find $\omega$ from SHM eqn. |
| $= T(1/4 - 1/6)$, $T = 12$ [s] | A1 | |
| $\omega^2(a^2 - x^2) = \frac{1}{4}v_{\max}^2$ | M1 | Find distance $x$ from O when speed is half max |
| $v_{\max} = a\omega$ [= $\pi a/6$] | M1 | Use maximum speed |
| $x^2 = \frac{3}{4}a^2$ | | note $x$ is independent of $\omega$ |
| $x = \sqrt{3}a/2$ or $0.866a$ | A1 | |
| $2.5 = \omega^{-1}\sin^{-1}(\sqrt{2}/a) - \omega^{-1}\sin^{-1}(-\frac{1}{2})$ | | EITHER: Find $a$ from $x = a\sin\omega t$ |
| or $1.5 = \omega^{-1}\sin^{-1}(\sqrt{2}/a)$ | | |
| $\sin^{-1}(\sqrt{2}/a) = \pi/4$, $a = 2$ [m] | M1 A1 | |
| $2.5 = \omega^{-1}\cos^{-1}(-\sqrt{2}/a) - \omega^{-1}\cos^{-1}\frac{1}{2}$ | | OR: Find $a$ from $x = \cos\omega t$ |
| or $1.5 = \omega^{-1}\cos^{-1}(-\sqrt{2}/a) - \omega^{-1}\cos^{-1}0$ | | |
| $\cos^{-1}(-\sqrt{2}/a) = 3\pi/4$, $a = 2$ [m] | M1 A1 | |

**Part marks: 4, 3, 2 | Total: 9**

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3 A particle moves on a straight line $A O B$ in simple harmonic motion, where $A B = 2 a \mathrm {~m}$. The centre of the motion is $O$ and the particle is instantaneously at rest at $A$ and $B$. The point $M$ is the mid-point of $O B$. The particle passes through $M$ moving towards $O$ and next achieves its maximum speed one second later. Find the period of the motion.

Find the distance of the particle from $O$ when its speed is equal to one half of its maximum speed.

At an instant 2.5 seconds after the particle passes through $M$ moving towards $O$, the distance of the particle from $O$ is $\sqrt { } 2 \mathrm {~m}$. Find, in metres, the amplitude of the motion.

\hfill \mbox{\textit{CAIE FP2 2015 Q3 [9]}}
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Q1 6 Q2 7 Q3 9 Q4 10 Q5 11 Q6 8 Q7 11 Q8 12 Q9 12 Q10 EITHER Q10 OR