Edexcel M3 2016 June — Question 2 9 marks

Exam BoardEdexcel
ModuleM3 (Mechanics 3)
Year2016
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSimple Harmonic Motion
TypeProve SHM and find period: vertical spring/string (single attachment)
DifficultyStandard +0.3 This is a standard SHM question requiring routine application of Hooke's law for equilibrium, then showing the restoring force is proportional to displacement (F = -kx form), and finally using the standard period formula. All steps follow textbook procedures with no novel insight required, making it slightly easier than average for M3.
Spec4.10f Simple harmonic motion: x'' = -omega^2 x6.02h Elastic PE: 1/2 k x^2

A light elastic spring, of natural length \(5a\) and modulus of elasticity \(10mg\), has one end attached to a fixed point \(A\) on a ceiling. A particle \(P\) of mass \(2m\) is attached to the other end of the spring and \(P\) hangs freely in equilibrium at the point \(O\).
  1. Find the distance \(AO\). [3]
The particle is now pulled vertically downwards a distance \(\frac{1}{2}a\) from \(O\) and released from rest.
  1. Show that \(P\) moves with simple harmonic motion. [4]
  2. Find the period of the motion. [2]

Part (a)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(T = 2mg = \frac{10mgx}{5a}\)M1 Resolve vertically using HL for the tension
\(x = a\)A1 Obtaining \(x = a\)
\(AO = 5a + a = 6a\)A1ft (3) Add their extension to \(5a\)
Part (b)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(2mg - T = 2m\ddot{x}\) or \(2ma\)M1 Forming an equation of motion, acceleration \(\ddot{x}\) or \(a\)
\(2mg - \frac{10mg(a + x)}{5a} = 2m\ddot{x}\)M1A1 Use HL to express the tension in terms of \(x\), (centre not nec at the correct point), acceleration \(\ddot{x}\) or \(a\)
\(\ddot{x} = -\frac{gx}{a}\) \(\therefore\) SHMA1cso (4) Simplify to \(\ddot{x} = -\frac{gx}{a}\) and state SHM
All marks available for an algebraic solution (ie no sub made for any/all of mass, \(\lambda\), \(l\))
Part (c)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Period \(= \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{a}{g}}\) oeM1A1 (2) Using period \(= \frac{2\pi}{\omega}\) with their \(\omega\) from a "SHM" equation. (ie \(\ddot{x} = a = \pm\omega^2 x\)) [9]
Key Notes:
- (a)M1: Resolve vertically using HL for the tension
- (a)A1: Obtaining \(x = a\)
- (a)A1ft: Add their extension to \(5a\)
- (b)M1: Forming an equation of motion, acceleration \(\ddot{x}\) or \(a\)
- (b)M1: Use HL to express the tension in terms of \(x\), (centre not nec at the correct point), acceleration \(\ddot{x}\) or \(a\)
- (b)A1cso: Simplify to \(\ddot{x} = -\frac{gx}{a}\) and state SHM
- (c)M1: Using period \(= \frac{2\pi}{\omega}\) with their \(\omega\) from a "SHM" equation
- (c)A1cso: Correct period, any equivalent form but not fractions within fractions. \(\omega\) must have come from a correct equation and substitutions for mass, \(\lambda\), \(l\) made.
### Part (a)

| Answer/Working | Marks | Guidance |
|---|---|---|
| $T = 2mg = \frac{10mgx}{5a}$ | M1 | Resolve vertically using HL for the tension |
| $x = a$ | A1 | Obtaining $x = a$ |
| $AO = 5a + a = 6a$ | A1ft | (3) Add their extension to $5a$ |

**Part (b)**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $2mg - T = 2m\ddot{x}$ or $2ma$ | M1 | Forming an equation of motion, acceleration $\ddot{x}$ or $a$ |
| $2mg - \frac{10mg(a + x)}{5a} = 2m\ddot{x}$ | M1A1 | Use HL to express the tension in terms of $x$, (centre not nec at the correct point), acceleration $\ddot{x}$ or $a$ |
| $\ddot{x} = -\frac{gx}{a}$ $\therefore$ SHM | A1cso | (4) Simplify to $\ddot{x} = -\frac{gx}{a}$ and state SHM |

All marks available for an algebraic solution (ie no sub made for any/all of mass, $\lambda$, $l$)

**Part (c)**

| Answer/Working | Marks | Guidance |
|---|---|---|
| Period $= \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{a}{g}}$ oe | M1A1 | (2) Using period $= \frac{2\pi}{\omega}$ with their $\omega$ from a "SHM" equation. (ie $\ddot{x} = a = \pm\omega^2 x$) [9] |

**Key Notes:**
- (a)M1: Resolve vertically using HL for the tension
- (a)A1: Obtaining $x = a$
- (a)A1ft: Add their extension to $5a$
- (b)M1: Forming an equation of motion, acceleration $\ddot{x}$ or $a$
- (b)M1: Use HL to express the tension in terms of $x$, (centre not nec at the correct point), acceleration $\ddot{x}$ or $a$
- (b)A1cso: Simplify to $\ddot{x} = -\frac{gx}{a}$ and state SHM
- (c)M1: Using period $= \frac{2\pi}{\omega}$ with their $\omega$ from a "SHM" equation
- (c)A1cso: Correct period, any equivalent form but not fractions within fractions. $\omega$ must have come from a correct equation and substitutions for mass, $\lambda$, $l$ made.

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A light elastic spring, of natural length $5a$ and modulus of elasticity $10mg$, has one end attached to a fixed point $A$ on a ceiling. A particle $P$ of mass $2m$ is attached to the other end of the spring and $P$ hangs freely in equilibrium at the point $O$.

\begin{enumerate}[label=(\alph*)]
\item Find the distance $AO$. [3]
\end{enumerate}

The particle is now pulled vertically downwards a distance $\frac{1}{2}a$ from $O$ and released from rest.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Show that $P$ moves with simple harmonic motion. [4]
\item Find the period of the motion. [2]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M3 2016 Q2 [9]}}