CAIE FP2 2013 November — Question 4 10 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2013
SessionNovember
Marks10
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TopicSimple Harmonic Motion
TypeSpeed at given displacement
DifficultyChallenging +1.2 This is a standard SHM problem requiring students to establish SHM from Hooke's law, find the period using ω, and use the energy equation v² = ω²(a² - x²) to find positions at a given speed. While it involves multiple steps and requires understanding of elastic strings in vertical motion, the techniques are well-practiced in Further Maths SHM topics with no novel insight required.
Spec6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle
4 A particle \(P\) of mass \(m\) is attached to one end of a light elastic string of natural length 4a. The other end of the string is attached to a fixed point \(O\). The particle rests in equilibrium at the point \(E\), vertically below \(O\), where \(O E = 5 a\). The particle is pulled down a vertical distance \(\frac { 1 } { 2 } a\) from \(E\) and released from rest. Show that the motion of \(P\) is simple harmonic and state the period of the motion. Find the two possible values of the distance \(O P\) when the speed of \(P\) is equal to one half of its maximum speed.
Question 4:
AnswerMarks Guidance
Working/AnswerMarks Guidance
\(\lambda a / 4a = mg\ [\lambda = 4mg]\)M1 A1 Resolve vertically at equilibrium with modulus \(\lambda\)
\(m\,d^2x/dt^2 = mg - \lambda(a+x)/4a\)— Newton's Law at general point
\([or\ -mg + \lambda(a-x)/4a]\)M1 A1
\(d^2x/dt^2 = -(g/a)\,x\)A1 (B1) Simplify to give standard SHM eqn. S.R.: Stating without derivation (max 4/6)
\(T = 2\pi/\sqrt{(g/a)}\) or \(2\pi\sqrt{(a/g)}\)B1 Find period \(T = 2\pi/\omega\) with \(\omega = \sqrt{(g/a)}\)
\(\omega^2(A^2 - x^2) = \frac{1}{4}\omega^2 A^2\)M1 A1 Equate speed at \(P\) to one-half maximum speed
\(x^2 = \frac{3}{4}A^2 = \frac{3}{4}(\frac{1}{2}a)^2\ [= 3a^2/16]\)A1 Find \(x^2\)
\(OP = (5 \pm \frac{1}{4}\sqrt{3})\,a\) (A.E.F.)A1 Find \(OP\)
Part totals: 6, 4Total: 10 
## Question 4:

| Working/Answer | Marks | Guidance |
|---|---|---|
| $\lambda a / 4a = mg\ [\lambda = 4mg]$ | M1 A1 | Resolve vertically at equilibrium with modulus $\lambda$ |
| $m\,d^2x/dt^2 = mg - \lambda(a+x)/4a$ | — | Newton's Law at general point |
| $[or\ -mg + \lambda(a-x)/4a]$ | M1 A1 | |
| $d^2x/dt^2 = -(g/a)\,x$ | A1 (B1) | Simplify to give standard SHM eqn. S.R.: Stating without derivation (max 4/6) |
| $T = 2\pi/\sqrt{(g/a)}$ or $2\pi\sqrt{(a/g)}$ | B1 | Find period $T = 2\pi/\omega$ with $\omega = \sqrt{(g/a)}$ |
| $\omega^2(A^2 - x^2) = \frac{1}{4}\omega^2 A^2$ | M1 A1 | Equate speed at $P$ to one-half maximum speed |
| $x^2 = \frac{3}{4}A^2 = \frac{3}{4}(\frac{1}{2}a)^2\ [= 3a^2/16]$ | A1 | Find $x^2$ |
| $OP = (5 \pm \frac{1}{4}\sqrt{3})\,a$ (A.E.F.) | A1 | Find $OP$ |

**Part totals: 6, 4 | Total: 10**

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4 A particle $P$ of mass $m$ is attached to one end of a light elastic string of natural length 4a. The other end of the string is attached to a fixed point $O$. The particle rests in equilibrium at the point $E$, vertically below $O$, where $O E = 5 a$. The particle is pulled down a vertical distance $\frac { 1 } { 2 } a$ from $E$ and released from rest. Show that the motion of $P$ is simple harmonic and state the period of the motion.

Find the two possible values of the distance $O P$ when the speed of $P$ is equal to one half of its maximum speed.

\hfill \mbox{\textit{CAIE FP2 2013 Q4 [10]}}
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