| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2016 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | SHM on inclined plane |
| Difficulty | Challenging +1.2 This is a standard M3 SHM question requiring proof of SHM condition, finding equilibrium position, and calculating period. While it involves multiple parts and requires careful application of Hooke's law and SHM formulas, the techniques are routine for Further Maths students. The multi-step nature and need to track when the string becomes slack elevates it slightly above average difficulty, but it follows a predictable template for elastic string SHM problems. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(T = mg\sin\alpha = \frac{3}{5}mg\) | B1 | \(T = \frac{3}{5}mg\) shown explicitly or used |
| \(T = \frac{\lambda \times \frac{1}{5}l}{l} = \frac{3}{5}mg\) | M1 | Attempt Hooke's law |
| \(\lambda = 3mg\) * | A1cso (3) | Obtain given result |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(mg\sin\alpha - T = m\ddot{x}\) | ||
| \(\frac{3}{5}mg - \frac{3mg(\frac{1}{5}l + x)}{l} = m\ddot{x}\) | M1A1 | Equation of motion along plane inc \(T\) in terms of \(l\) and \(x\); correct equation |
| \(\ddot{x} = -\frac{3g}{l}x\) | dM1 | Re-arrange to required form; acceleration must be \(\ddot{x}\) |
| \(\therefore\) SHM | A1 (4) | Correct result; SHM stated |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \( | \ddot{x} | _{\max} = a\omega^2 = \frac{1}{2}l \times \frac{3g}{l} = \frac{3g}{2}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Time from \(D\) to \(B\): \(\frac{1}{4}l = \frac{1}{2}l\sin\sqrt{\frac{3g}{l}}t_1\) | M1 | Find time from \(D\) to \(B\) or from \(C\) to \(D\) |
| \(t_1 = \frac{\pi}{6}\sqrt{\frac{l}{3g}}\) | ||
| Time from \(B\) to nat. length: \(\frac{1}{5}l = \frac{1}{2}l\sin\sqrt{\frac{3g}{l}}t_2\) | M1 | Find time from \(B\) to natural length or from \(C\) to natural length |
| \(t_2 = \sqrt{\frac{l}{3g}}\sin^{-1}\frac{2}{5}\) | ||
| Total time \(= \left(\frac{\pi}{6} + \sin^{-1}\frac{2}{5}\right)\sqrt{\frac{l}{3g}} = 0.54\sqrt{\frac{l}{g}}\), \(k=0.54\) | ddM1,A1cao (4) [13] | Add/subtract the two times; correct result for \(k\), must be 2 sf |
## Question 5:
### Part (a):
| Working | Marks | Guidance |
|---------|-------|----------|
| $T = mg\sin\alpha = \frac{3}{5}mg$ | B1 | $T = \frac{3}{5}mg$ shown explicitly or used |
| $T = \frac{\lambda \times \frac{1}{5}l}{l} = \frac{3}{5}mg$ | M1 | Attempt Hooke's law |
| $\lambda = 3mg$ * | A1cso (3) | Obtain given result |
### Part (b):
| Working | Marks | Guidance |
|---------|-------|----------|
| $mg\sin\alpha - T = m\ddot{x}$ | | |
| $\frac{3}{5}mg - \frac{3mg(\frac{1}{5}l + x)}{l} = m\ddot{x}$ | M1A1 | Equation of motion along plane inc $T$ in terms of $l$ and $x$; correct equation |
| $\ddot{x} = -\frac{3g}{l}x$ | dM1 | Re-arrange to required form; acceleration must be $\ddot{x}$ |
| $\therefore$ SHM | A1 (4) | Correct result; SHM stated |
### Part (c):
| Working | Marks | Guidance |
|---------|-------|----------|
| $|\ddot{x}|_{\max} = a\omega^2 = \frac{1}{2}l \times \frac{3g}{l} = \frac{3g}{2}$ | M1A1ft (2) | Use $|\ddot{x}|_{\max} = a\omega^2$; correct answer following through $\omega$ |
### Part (d):
| Working | Marks | Guidance |
|---------|-------|----------|
| Time from $D$ to $B$: $\frac{1}{4}l = \frac{1}{2}l\sin\sqrt{\frac{3g}{l}}t_1$ | M1 | Find time from $D$ to $B$ or from $C$ to $D$ |
| $t_1 = \frac{\pi}{6}\sqrt{\frac{l}{3g}}$ | | |
| Time from $B$ to nat. length: $\frac{1}{5}l = \frac{1}{2}l\sin\sqrt{\frac{3g}{l}}t_2$ | M1 | Find time from $B$ to natural length or from $C$ to natural length |
| $t_2 = \sqrt{\frac{l}{3g}}\sin^{-1}\frac{2}{5}$ | | |
| Total time $= \left(\frac{\pi}{6} + \sin^{-1}\frac{2}{5}\right)\sqrt{\frac{l}{3g}} = 0.54\sqrt{\frac{l}{g}}$, $k=0.54$ | ddM1,A1cao (4) [13] | Add/subtract the two times; correct result for $k$, must be 2 sf |
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ffe0bc72-3136-48d9-9d5b-4a364d134070-07_371_800_262_573}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A particle $P$ of mass $m$ is attached to one end of a light elastic string, of natural length $l$ and modulus of elasticity $\lambda$. The other end of the string is attached to a fixed point $A$ on a smooth plane inclined at angle $\alpha$ to the horizontal, where $\sin \alpha = \frac { 3 } { 5 }$. The particle rests in equilibrium on the plane at the point $B$ with the string lying along a line of greatest slope of the plane, as shown in Figure 2.
Given that $A B = \frac { 6 } { 5 } l$
\begin{enumerate}[label=(\alph*)]
\item show that $\lambda = 3 \mathrm { mg }$
The particle is pulled down the line of greatest slope to the point $C$, where $B C = \frac { 1 } { 2 } l$, and released from rest.
\item Show that, while the string remains taut, $P$ moves with simple harmonic motion about centre $B$.
\item Find the greatest magnitude of the acceleration of $P$ while the string remains taut.
The point $D$ is the midpoint of $B C$. The time taken by $P$ to move directly from $D$ to the point where the string becomes slack for the first time is $k \sqrt { \frac { l } { g } }$, where $k$ is a constant.
\item Find, to 2 significant figures, the value of $k$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2016 Q5 [13]}}