Pre-U Pre-U 9795/2 2015 June — Question 8 4 marks

Exam BoardPre-U
ModulePre-U 9795/2 (Pre-U Further Mathematics Paper 2)
Year2015
SessionJune
Marks4
TopicSimple Harmonic Motion
TypeTime to travel between positions
DifficultyStandard +0.3 This is a standard SHM question requiring identification of equilibrium position, verification of SHM equation (F = -kx), calculation of period using ω = √(λ/ml), and finding time between positions using the standard SHM formula x = a cos(ωt). All steps are routine applications of well-practiced techniques with no novel insight required, making it slightly easier than average.
Spec4.10f Simple harmonic motion: x'' = -omega^2 x6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle

8 \includegraphics[max width=\textwidth, alt={}, center]{86cc07e7-ea69-4480-96c8-82b818445199-4_182_803_264_671} A light spring of modulus of elasticity 8 N and natural length 0.4 m has one end fixed to a smooth horizontal table at a fixed point \(L\). A particle of mass 0.2 kg is attached to the other end of the spring and pulled out horizontally to a point \(M\) on the table, so that the spring is extended by 0.2 m . The particle is then released from rest. The mid-point of \(L M\) is \(N\) and the point \(O\) is on \(L M\) such that \(L O = 0.4 \mathrm {~m}\) (see diagram).
  1. Show that the particle moves in simple harmonic motion with centre \(O\) and state the exact period of its motion.
  2. Find the exact time taken for the particle to move directly from \(M\) to \(N\).

Question 8(i) and 8(ii)
(i)
Hooke's law: \(T = \dfrac{8x}{0.4} = 20x\) — B1 (Can be specific \(x\))
Newton II: \(\dfrac{1}{5}\ddot{x} = -20x\) — M1 (Needs general \(x\), \(-\) sign)
\(\Rightarrow \ddot{x} = -100x\) which is SHM — A1 (A.e. exact f.)
Period \(= \dfrac{2\pi}{10} = \dfrac{1}{5}\pi\) seconds — A1
[4]
(ii)
\(-0.1 = 0.2\cos 10t\) — M1A1 (Or: Obtain \(\dfrac{1}{60}\pi\))
\(\Rightarrow 10t = \dfrac{2\pi}{3} \Rightarrow t = \dfrac{1}{15}\pi\) seconds — M1A1 (Add quarter period to get \(\dfrac{1}{15}\pi\))
[4]
**Question 8(i) and 8(ii)**

**(i)**
Hooke's law: $T = \dfrac{8x}{0.4} = 20x$ — B1 (Can be specific $x$)

Newton II: $\dfrac{1}{5}\ddot{x} = -20x$ — M1 (Needs general $x$, $-$ sign)

$\Rightarrow \ddot{x} = -100x$ which is SHM — A1 (A.e. exact f.)

Period $= \dfrac{2\pi}{10} = \dfrac{1}{5}\pi$ seconds — A1

**[4]**

**(ii)**
$-0.1 = 0.2\cos 10t$ — M1A1 (Or: Obtain $\dfrac{1}{60}\pi$)

$\Rightarrow 10t = \dfrac{2\pi}{3} \Rightarrow t = \dfrac{1}{15}\pi$ seconds — M1A1 (Add quarter period to get $\dfrac{1}{15}\pi$)

**[4]**
8\\
\includegraphics[max width=\textwidth, alt={}, center]{86cc07e7-ea69-4480-96c8-82b818445199-4_182_803_264_671}

A light spring of modulus of elasticity 8 N and natural length 0.4 m has one end fixed to a smooth horizontal table at a fixed point $L$. A particle of mass 0.2 kg is attached to the other end of the spring and pulled out horizontally to a point $M$ on the table, so that the spring is extended by 0.2 m . The particle is then released from rest. The mid-point of $L M$ is $N$ and the point $O$ is on $L M$ such that $L O = 0.4 \mathrm {~m}$ (see diagram).\\
(i) Show that the particle moves in simple harmonic motion with centre $O$ and state the exact period of its motion.\\
(ii) Find the exact time taken for the particle to move directly from $M$ to $N$.

\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2015 Q8 [4]}}