| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2004 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Prove SHM and find period: horizontal or non-standard geometry |
| Difficulty | Standard +0.3 This is a standard SHM question requiring application of Hooke's law to establish the SHM equation, then using standard formulas for period and speed. Part (c) involves finding time intervals using inverse trig, which is routine for M3. All steps follow textbook methods with no novel insight required, making it slightly easier than average. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x4.10g Damped oscillations: model and interpret6.02g Hooke's law: T = k*x or T = lambda*x/l6.02h Elastic PE: 1/2 k x^2 |
| Answer | Marks |
|---|---|
| (a) Applying Hooke's Law correctly: e.g. \(T = \frac{48x}{0.6}\) | M1 |
| Equation of motion: \((-) T = 0.2\ddot{x}\) | M1 |
| Correct equation of motion: e.g. \(-\frac{48x}{0.6} = 0.2\ddot{x}\) | A1 |
| Writing in form \(\ddot{x} = -\omega^2x\), and stating motion is SHM | A1 ⋈ |
| Period \(= \frac{2\pi}{\omega} = \frac{2\pi}{20} = \frac{\pi}{10}\) * | A1 (5) |
| Answer | Marks |
|---|---|
| (b) \(\max v = a\omega\) with values substituted; \(= 0.3 \times 20 = 6\) ms\(^{-1}\) | M1A1(2) |
| (c) Using \(x = 0.3\cos 20t\) or \(x = 0.3\sin 20t\) | M1 |
| Using \(x = 0.15\) to give either \(\cos 20t = \frac{1}{2}\) or \(\sin 20t = \frac{1}{2}\) | M1 |
| Either \(t = \frac{\pi}{60}, \frac{5\pi}{60}\) or \(T = \frac{\pi}{120}\) | A1 |
| Complete method for time: | M1 |
| \(t_2 - t_1\), or \(\frac{\pi}{10} - 2t_1\), or \(2\left(\frac{\pi}{40} + T\right)\) | |
| Time \(= \frac{\pi}{15}\) s (must be in terms of \(\pi\)) | A1 (5) |
| [12] |
(a) Applying Hooke's Law correctly: e.g. $T = \frac{48x}{0.6}$ | M1 |
Equation of motion: $(-) T = 0.2\ddot{x}$ | M1 |
Correct equation of motion: e.g. $-\frac{48x}{0.6} = 0.2\ddot{x}$ | A1 |
Writing in form $\ddot{x} = -\omega^2x$, and stating motion is SHM | A1 ⋈ |
Period $= \frac{2\pi}{\omega} = \frac{2\pi}{20} = \frac{\pi}{10}$ * | A1 (5) |
(no incorrect working seen)
[If measure $x$ from B or A, final 2 marks only available if equation of motion is reduced to $\ddot{X} = -\omega^2X$]
(b) $\max v = a\omega$ with values substituted; $= 0.3 \times 20 = 6$ ms$^{-1}$ | M1A1(2) |
(c) Using $x = 0.3\cos 20t$ or $x = 0.3\sin 20t$ | M1 |
Using $x = 0.15$ to give either $\cos 20t = \frac{1}{2}$ or $\sin 20t = \frac{1}{2}$ | M1 |
Either $t = \frac{\pi}{60}, \frac{5\pi}{60}$ or $T = \frac{\pi}{120}$ | A1 |
Complete method for time: | M1 |
$t_2 - t_1$, or $\frac{\pi}{10} - 2t_1$, or $2\left(\frac{\pi}{40} + T\right)$ | |
Time $= \frac{\pi}{15}$ s (must be in terms of $\pi$) | A1 (5) |
| [12] |
---5. A piston in a machine is modelled as a particle of mass 0.2 kg attached to one end $A$ of a light elastic spring, of natural length 0.6 m and modulus of elasticity 48 N . The other end $B$ of the spring is fixed and the piston is free to move in a horizontal tube which is assumed to be smooth. The piston is released from rest when $A B = 0.9 \mathrm {~m}$.
\begin{enumerate}[label=(\alph*)]
\item Prove that the motion of the piston is simple harmonic with period $\frac { \pi } { 10 } \mathrm {~s}$.\\
(5)
\item Find the maximum speed of the piston.\\
(2)
\item Find, in terms of $\pi$, the length of time during each oscillation for which the length of the spring is less than 0.75 m .\\
(5)
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2004 Q5 [12]}}