| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2015 |
| Session | January |
| Marks | 16 |
| 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 with familiar structure: equilibrium to find modulus, show SHM with given period, find maximum acceleration, and calculate time with slack string. While it requires multiple techniques (Hooke's law, SHM equations, energy methods), each part follows predictable patterns that M3 students practice extensively. The inclined plane adds mild complexity but the 30° angle and clean fractions keep calculations manageable. Slightly above average due to the multi-part nature and part (d) requiring careful consideration of when the string goes slack. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(T = \frac{\lambda a/5}{a}\) | M1A1 | M1 Hooke's Law used to find \(T\) at \(B\); A1 correct equation |
| \(T = mg\cos 60 = \frac{1}{2}mg\) | ||
| \(\frac{1}{2}mg = \frac{\lambda}{5}\ \Rightarrow\ \lambda = \frac{5}{2}mg\) | M1A1 (4) | M1 eliminating \(T\) by resolving along plane; A1cso correct value for \(\lambda\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| When string has length \(\left(\frac{6a}{5}+x\right)\): \(\frac{1}{2}mg - \frac{5}{2}mg\left(\frac{a/5+x}{a}\right) = m\ddot{x}\) | M1A1A1 | M1 NL2 along plane when extension is \(\frac{a}{5}+x\); A1A1 deduct one per error (difference of forces wrong way is one error only); mass \(\times\) acceleration (not \(\ddot{x}\)) is also an error |
| \(-\frac{5g}{2a}x = \ddot{x}\ \Rightarrow\) SHM | DM1A1 | DM1 simplify to correct form, acceleration must be \(\ddot{x}\); A1cso correct final equation AND conclusion |
| Period \(= 2\pi\sqrt{\frac{2a}{5g}}\) | A1 (6) | A1 correct period |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Max accel \(= \omega^2 \times \text{amp} = \omega^2\frac{a}{5} = \frac{5g}{2a} \times \frac{a}{5} = \frac{g}{2}\) | M1A1 (2) | M1 obtaining max acceleration, amp \(\neq a\); A1 correct max acceleration (no ft) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x = \frac{a}{5}\sin\omega t\) | ||
| \(\frac{a}{10} = \frac{a}{5}\sin\omega t\) | M1 | M1 using equation for \(x\) - sin or cos form and solving for \(t\); must use radians and \(\omega = \sqrt{\frac{5g}{2a}}\), amp \(\neq a\) |
| \(\omega t = \sin^{-1}0.5 = \frac{\pi}{6}\) | ||
| \(t = \frac{\pi}{6\omega} = \frac{\pi}{6}\sqrt{\frac{2a}{5g}}\) | A1 | A1 correct value for \(t\) from their equation |
| Total time \(= \frac{\pi}{6}\sqrt{\frac{2a}{5g}} + \frac{\pi}{2}\sqrt{\frac{2a}{5g}} = \frac{2\pi}{3}\sqrt{\frac{2a}{5g}}\) | M1A1 (4) | M1 complete to obtain required time; A1 correct total time |
| If time from end point to \(x = -\frac{a}{10}\) is found mark M1M1A1A1 |
# Question 7:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $T = \frac{\lambda a/5}{a}$ | M1A1 | M1 Hooke's Law used to find $T$ at $B$; A1 correct equation |
| $T = mg\cos 60 = \frac{1}{2}mg$ | | |
| $\frac{1}{2}mg = \frac{\lambda}{5}\ \Rightarrow\ \lambda = \frac{5}{2}mg$ | M1A1 (4) | M1 eliminating $T$ by resolving along plane; A1cso correct value for $\lambda$ |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| When string has length $\left(\frac{6a}{5}+x\right)$: $\frac{1}{2}mg - \frac{5}{2}mg\left(\frac{a/5+x}{a}\right) = m\ddot{x}$ | M1A1A1 | M1 NL2 along plane when extension is $\frac{a}{5}+x$; A1A1 deduct one per error (difference of forces wrong way is one error only); mass $\times$ acceleration (not $\ddot{x}$) is also an error |
| $-\frac{5g}{2a}x = \ddot{x}\ \Rightarrow$ SHM | DM1A1 | DM1 simplify to correct form, acceleration must be $\ddot{x}$; A1cso correct final equation AND conclusion |
| Period $= 2\pi\sqrt{\frac{2a}{5g}}$ | A1 (6) | A1 correct period |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Max accel $= \omega^2 \times \text{amp} = \omega^2\frac{a}{5} = \frac{5g}{2a} \times \frac{a}{5} = \frac{g}{2}$ | M1A1 (2) | M1 obtaining max acceleration, amp $\neq a$; A1 correct max acceleration (no ft) |
## Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x = \frac{a}{5}\sin\omega t$ | | |
| $\frac{a}{10} = \frac{a}{5}\sin\omega t$ | M1 | M1 using equation for $x$ - sin or cos form and solving for $t$; must use radians and $\omega = \sqrt{\frac{5g}{2a}}$, amp $\neq a$ |
| $\omega t = \sin^{-1}0.5 = \frac{\pi}{6}$ | | |
| $t = \frac{\pi}{6\omega} = \frac{\pi}{6}\sqrt{\frac{2a}{5g}}$ | A1 | A1 correct value for $t$ from their equation |
| Total time $= \frac{\pi}{6}\sqrt{\frac{2a}{5g}} + \frac{\pi}{2}\sqrt{\frac{2a}{5g}} = \frac{2\pi}{3}\sqrt{\frac{2a}{5g}}$ | M1A1 (4) | M1 complete to obtain required time; A1 correct total time |
| If time from end point to $x = -\frac{a}{10}$ is found mark M1M1A1A1 | | |7. A particle $P$ of mass $m$ is attached to one end of a light elastic string, of natural length $a$ and modulus of elasticity $\lambda$. The other end of the string is attached to a fixed point $A$ on a smooth plane which is inclined at $30 ^ { \circ }$ to the horizontal. The string lies along a line of greatest slope of the plane. The particle rests in equilibrium at the point $B$, where $B$ is lower than $A$ and $A B = \frac { 6 } { 5 } a$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\lambda = \frac { 5 } { 2 } m g$.
The particle is now pulled down a line of greatest slope to the point $C$, where $B C = \frac { 1 } { 5 } a$, and released from rest.
\item Show that $P$ moves with simple harmonic motion of period $2 \pi \sqrt { \frac { 2 a } { 5 g } }$
\item Find, in terms of $g$, the greatest magnitude of the acceleration of $P$ while the string is taut.
The midpoint of $B C$ is $D$ and the string becomes slack for the first time at the point $E$.
\item Find, in terms of $a$ and $g$, the time taken by $P$ to travel directly from $D$ to $E$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2015 Q7 [16]}}