| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2005 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Maximum speed in SHM |
| Difficulty | Standard +0.3 This is a standard M3 SHM question requiring energy conservation (part a), verification of SHM conditions using Hooke's law (part b), and application of standard SHM formulas for period and maximum speed (part c). While it involves multiple steps and careful bookkeeping of extensions, it follows predictable patterns taught in M3 with no novel insights required—slightly easier than average due to its structured, methodical nature. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x6.02g Hooke's law: T = k*x or T = lambda*x/l6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Content | Marks | Notes |
| KE Loss + PE Loss \(= \triangle\)PE Gain | M1 | |
| \(\frac{1}{2} \cdot m(2g)L + mgL \cdot 3L = \frac{\lambda(3L)^2}{2L}\) | M1 A2 | G1(c4) |
| \(\star\) \(\frac{3mg}{8} = \lambda\) | A1 | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Content | Marks | Notes |
| \(mg - T = m\ddot{x}\) | M1 | |
| \(m\frac{3mg}{8L}(x+\ell) = m\ddot{x}\) | A1 | |
| \(-\frac{3g}{8L} \cdot x = \ddot{x}\) | A1 esc. | (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Content | Marks | Notes |
| (i) Period \(= \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{8L}{3g}} = 3\pi\sqrt{\frac{L}{2g}}\) | M1 | |
| (ii) \(m_5 = \frac{3mg}{8L} \cdot e \Rightarrow e = \frac{12}{8}\) | B1 | |
| \(a = 3L - 4/8 = \frac{15L}{8}\) | M1 | |
| \(v_{\max} = a\omega = \frac{15L}{8} \cdot \sqrt{\frac{3g}{8L}} = \frac{5\sqrt{2gL}}{4}\) | M1 A1 | (5), (14) |
## Part (a)
| Content | Marks | Notes |
|---------|-------|-------|
| KE Loss + PE Loss $= \triangle$PE Gain | M1 | |
| $\frac{1}{2} \cdot m(2g)L + mgL \cdot 3L = \frac{\lambda(3L)^2}{2L}$ | M1 A2 | G1(c4) |
| $\star$ $\frac{3mg}{8} = \lambda$ | A1 | (4) |
## Part (b)
| Content | Marks | Notes |
|---------|-------|-------|
| $mg - T = m\ddot{x}$ | M1 | |
| $m\frac{3mg}{8L}(x+\ell) = m\ddot{x}$ | A1 | |
| $-\frac{3g}{8L} \cdot x = \ddot{x}$ | A1 esc. | (5) |
## Part (c)
| Content | Marks | Notes |
|---------|-------|-------|
| (i) Period $= \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{8L}{3g}} = 3\pi\sqrt{\frac{L}{2g}}$ | M1 | |
| (ii) $m_5 = \frac{3mg}{8L} \cdot e \Rightarrow e = \frac{12}{8}$ | B1 | |
| $a = 3L - 4/8 = \frac{15L}{8}$ | M1 | |
| $v_{\max} = a\omega = \frac{15L}{8} \cdot \sqrt{\frac{3g}{8L}} = \frac{5\sqrt{2gL}}{4}$ | M1 A1 | (5), (14) |
---A light spring of natural length $L$ has one end attached to a fixed point $A$. A particle $P$ of mass $m$ is attached to the other end of the spring. The particle is moving vertically. As it passes through the point $B$ below $A$, where $AB = L$, its speed is $\sqrt{(2gL)}$. The particle comes to instantaneous rest at a point $C$, $4L$ below $A$.
\begin{enumerate}[label=(\alph*)]
\item Show that the modulus of elasticity of the spring is $\frac{8mg}{9}$. [4]
\end{enumerate}
At the point $D$ the tension in the spring is $mg$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Show that $P$ performs simple harmonic motion with centre $D$. [5]
\item Find, in terms of $L$ and $g$,
\begin{enumerate}[label=(\roman*)]
\item the period of the simple harmonic motion,
\item the maximum speed of $P$.
\end{enumerate} [5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2005 Q6 [14]}}