| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2017 |
| Session | January |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Time to travel between positions |
| Difficulty | Challenging +1.2 This is a standard M3/Further Mechanics SHM question requiring multiple techniques (equilibrium position, SHM verification, period formula, amplitude from max speed, and time calculation using arcsin). While it involves several parts and careful bookkeeping of the equilibrium position as SHM centre, each step follows predictable patterns taught in M3. Part (e) requires integrating or using the SHM time formula with careful attention to positions, which elevates it slightly above routine but remains within standard Further Maths territory. |
| Spec | 1.05a Sine, cosine, tangent: definitions for all arguments1.05o Trigonometric equations: solve in given intervals4.10f Simple harmonic motion: x'' = -omega^2 x6.02g Hooke's law: T = k*x or T = lambda*x/l6.02h Elastic PE: 1/2 k x^2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(2mg = \frac{20mge}{5l}\) | M1 | Resolve vertically using Hooke's law to find tension. Formula for HL to be correct |
| \(e = \frac{l}{2}\) | A1 | Solve equation to find equilibrium extension |
| \(AB = 5.5l\) | A1ft | Add \(5l\) to equilibrium extension to obtain \(AB\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(2mg - T = 2m\ddot{x}\) | M1 | Resolve vertically at general point. HL not needed; acceleration can be \(\ddot{x}\) or \(a\). Mass \(2m\) or \(m\) |
| \(2mg - \frac{20mg(0.5l+x)}{5l} = 2m\ddot{x}\) | dM1A1A1 | Use HL with extension \((0.5l+x)\) or \((e+x)\). Dependent on previous M mark. Correct difference of forces |
| \(\ddot{x} = -\frac{2g}{l}x \therefore\) SHM | A1 cso | Fully correct equation with acceleration \(\ddot{x}\). Correct equation starting \(\ddot{x}=\ldots\) and conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{l}{2g}} = \pi\sqrt{\frac{2l}{g}}\) | M1A1ft | Use Period \(= \frac{2\pi}{\omega}\) with their \(\omega\) from SHM equation. Correct period follow through their \(\omega\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(a\omega = \frac{1}{5}\sqrt{gl} = a\sqrt{\frac{2g}{l}}\) | M1A1ft | Use \(v = a\omega\) or \(v^2 = \omega^2(a^2-x^2)\) with \(x=0\), their \(\omega\) |
| \(a = \frac{l}{5\sqrt{2}}\) | A1 | Correct amplitude |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(BD = \frac{1}{2}a\) | ||
| \(-\frac{1}{2}a = a\cos\omega t\) | M1 | Use \(x = a\cos\omega t\) or \(x = a\sin\omega t\) with their \(\omega\), and \(x = \pm\frac{1}{2}a\) |
| \(\cos t\sqrt{\frac{2g}{l}} = -\frac{1}{2}\) | A1 | Correct equation (limited ft). Allow amplitude in terms of \(l\) if used but \(\omega\) correct |
| \(t = \sqrt{\frac{l}{2g}}\cos^{-1}\left(-\frac{1}{2}\right)\) | dM1 | Solve to \(t = \ldots\) Must be in radians. Complete correct method |
| \(t = \sqrt{\frac{l}{2g}} \times \frac{2\pi}{3} = \frac{2\pi}{3}\sqrt{\frac{l}{2g}}\) | A1cso | Correct time. cso applies to part (e) work only |
## Question 6:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $2mg = \frac{20mge}{5l}$ | M1 | Resolve vertically using Hooke's law to find tension. Formula for HL to be correct |
| $e = \frac{l}{2}$ | A1 | Solve equation to find equilibrium extension |
| $AB = 5.5l$ | A1ft | Add $5l$ to equilibrium extension to obtain $AB$ |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $2mg - T = 2m\ddot{x}$ | M1 | Resolve vertically at general point. HL not needed; acceleration can be $\ddot{x}$ or $a$. Mass $2m$ or $m$ |
| $2mg - \frac{20mg(0.5l+x)}{5l} = 2m\ddot{x}$ | dM1A1A1 | Use HL with extension $(0.5l+x)$ or $(e+x)$. Dependent on previous M mark. Correct difference of forces |
| $\ddot{x} = -\frac{2g}{l}x \therefore$ SHM | A1 cso | Fully correct equation with acceleration $\ddot{x}$. Correct equation starting $\ddot{x}=\ldots$ and conclusion |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{l}{2g}} = \pi\sqrt{\frac{2l}{g}}$ | M1A1ft | Use Period $= \frac{2\pi}{\omega}$ with their $\omega$ from SHM equation. Correct period follow through their $\omega$ |
### Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $a\omega = \frac{1}{5}\sqrt{gl} = a\sqrt{\frac{2g}{l}}$ | M1A1ft | Use $v = a\omega$ or $v^2 = \omega^2(a^2-x^2)$ with $x=0$, their $\omega$ |
| $a = \frac{l}{5\sqrt{2}}$ | A1 | Correct amplitude |
### Part (e):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $BD = \frac{1}{2}a$ | | |
| $-\frac{1}{2}a = a\cos\omega t$ | M1 | Use $x = a\cos\omega t$ or $x = a\sin\omega t$ with their $\omega$, and $x = \pm\frac{1}{2}a$ |
| $\cos t\sqrt{\frac{2g}{l}} = -\frac{1}{2}$ | A1 | Correct equation (limited ft). Allow amplitude in terms of $l$ if used but $\omega$ correct |
| $t = \sqrt{\frac{l}{2g}}\cos^{-1}\left(-\frac{1}{2}\right)$ | dM1 | Solve to $t = \ldots$ Must be in radians. Complete correct method |
| $t = \sqrt{\frac{l}{2g}} \times \frac{2\pi}{3} = \frac{2\pi}{3}\sqrt{\frac{l}{2g}}$ | A1cso | Correct time. cso applies to part (e) work only |
---6. One end of a light elastic string, of natural length 5 l and modulus of elasticity 20 mg , is attached to a fixed point $A$. A particle $P$ of mass $2 m$ is attached to the free end of the string and $P$ hangs freely in equilibrium at the point $B$.
\begin{enumerate}[label=(\alph*)]
\item Find the distance $A B$.\\
(3)
The particle is now pulled vertically downwards from $B$ to the point $C$ and released from rest. In the subsequent motion the string does not become slack.
\item Show that $P$ moves with simple harmonic motion with centre $B$.
\item Find the period of this motion.
The greatest speed of $P$ during this motion is $\frac { 1 } { 5 } \sqrt { g l }$
\item Find the amplitude of this motion.
The point $D$ is the midpoint of $B C$ and the point $E$ is the highest point reached by $P$.
\item Find the time taken by $P$ to move directly from $D$ to $E$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2017 Q6 [17]}}