| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2015 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Vertical SHM with two strings |
| Difficulty | Challenging +1.2 This is a multi-part M3 question requiring elastic string equilibrium analysis, EPE calculation, and SHM verification with period calculation. While it involves several mechanics concepts and multiple steps (6+7 marks), the techniques are standard for M3: applying Hooke's law to two strings, using energy methods, and showing SHM conditions. The 'show that' parts provide checkpoints, and the problem follows a familiar M3 template without requiring novel geometric insight or particularly complex algebra. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings |
| Answer | Marks |
|---|---|
| M1 | used at least once |
| A1 | |
| A1, M1 | CAO (AG); used at least once |
| A1 | |
| A1 [6] | may see 0.5 + 0.12 |
| Answer | Marks |
|---|---|
| M1* | allow sign errors, '\(m\)' wrong |
| A1 | |
| A1 | accept \(a = -64.89...x\), \(a = -\frac{2401}{37}x\) oe |
| A1 | dep on all first 3 marks |
| *M1 | must subsт for their \(\omega\) |
| A1 | allow if \(\omega\) correct |
| B1 [7] | justified at some point |
### Part (i)
**Answer:** use of $T = \frac{2\pi}{s}$
$T = \frac{10x0.2}{0.4} + \frac{12 \times 0.1}{0.5}$
$W = 7.4$ N
use of $E = \frac{2I}{2l}$
$E = \frac{10(0.2)^2}{2 \times 0.4} + \frac{12(0.1)^2}{2 \times 0.5}$
$E = 0.62$ (J)
| M1 | used at least once |
| A1 | |
| A1, M1 | CAO (AG); used at least once |
| A1 | |
| A1 [6] | may see 0.5 + 0.12 |
### Part (ii)
**Answer:** use of $F = ma$ when further extension is $x$
$7.4 - \frac{10 \times (x + 0.2)}{0.4} - \frac{12 \times (x + 0.1)}{0.5} = \frac{g}{a}$
$a = -\frac{49g}{7.4}$
SHM: $\omega^2 = \frac{49g}{7.4}$ (or $\frac{2401}{37}$ or 64.89189)
Use of $T = \frac{2\pi}{\omega}$
period is 0.780 (secs) $\frac{2\pi\sqrt{37}}{49}$
all subsequent motion is SHM because string does not become slack
| M1* | allow sign errors, '$m$' wrong |
| A1 | |
| A1 | accept $a = -64.89...x$, $a = -\frac{2401}{37}x$ oe |
| A1 | dep on all first 3 marks |
| *M1 | must subsт for their $\omega$ |
| A1 | allow if $\omega$ correct |
| | |
| B1 [7] | justified at some point |
**Guidance:** OR, when total length of string is $x$: $7.4 - \frac{10 \times (x - 0.4)}{0.4} - \frac{12 \times (x - 0.5)}{0.5} = \frac{g}{a}$
$a = -\frac{49g}{7.4}(x - 0.6)$
SHM about $x = 0.6$, and $\omega^2$ given
0.77998\includegraphics{figure_3}
A small object $P$ is attached to one end of each of two vertical light elastic strings. One string is of natural length $0.4$ m and has modulus of elasticity $10$ N; the other string is of natural length $0.5$ m and has modulus of elasticity $12$ N. The upper ends of both strings are attached to a fixed horizontal beam and $P$ hangs in equilibrium $0.6$ m below the beam (see diagram).
\begin{enumerate}[label=(\roman*)]
\item Show that the weight of $P$ is $7.4$ N and find the total elastic potential energy stored in the two strings when $P$ is hanging in equilibrium. [6]
\end{enumerate}
$P$ is then held at a point $0.7$ m below the beam with the strings vertical. $P$ is released from rest.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Show that, throughout the subsequent motion, $P$ performs simple harmonic motion, and find the period. [7]
\end{enumerate}
\hfill \mbox{\textit{OCR M3 2015 Q3 [13]}}