| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2021 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Vertical SHM with two strings |
| Difficulty | Challenging +1.2 This is a standard M3/Further Mechanics SHM question requiring equilibrium analysis with two strings, verification of SHM conditions using Hooke's law, and application of energy conservation. While it involves multiple strings and careful bookkeeping of extensions, the techniques are routine for this module—setting up force equations, showing restoring force proportional to displacement, and using standard SHM formulas. The multi-part structure and algebraic manipulation place it slightly above average difficulty, but it follows a predictable template for vertical SHM problems. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\dfrac{2mge_1}{2a}\) or \(\dfrac{6mg(4a-e_1)}{4a}\) | B1 | |
| \(mg + \dfrac{2mge_1}{2a} = \dfrac{6mg(4a-e_1)}{4a}\) | M1A1 | |
| Solve to find either extension | dM1 | |
| \(e_1 = 2a\) and \(e_2 = 4a - e_1 = 2a\)* | A1* |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(mg + \dfrac{2mge_1}{2a} = \dfrac{6mge_2}{4a}\), \(e_1 + e_2 = 4a\) | M1A1 | |
| Solve simultaneously | dM1 | |
| \(e_1 = 2a\) and \(e_2 = 2a\)* | A1* |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(mg + \dfrac{2mg(2a-x)}{2a} - \dfrac{6mg(2a+x)}{4a} = m\ddot{x}\) | M1A1A1 | |
| \(\ddot{x} = -\dfrac{5g}{2a}x \therefore\) SHM | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\omega^2 = \dfrac{5g}{2a}\) | B1ft | |
| \(v^2 = \dfrac{5g}{2a}\left(a^2 - \left(\dfrac{a}{2}\right)^2\right)\) | M1A1 | |
| \(v = \sqrt{\dfrac{15ga}{8}} = \dfrac{\sqrt{30ga}}{4}\) | A1 cso |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\dfrac{2mga^2}{4a}\) or \(\dfrac{6mg(3a)^2}{8a}\) or \(\dfrac{2mg\left(\frac{3a}{2}\right)^2}{4a}\) or \(\dfrac{6mg\left(\frac{5a}{2}\right)^2}{8a}\) | B1 | |
| \(\dfrac{2mga^2}{4a} + \dfrac{6mg(3a)^2}{8a} = \dfrac{2mg\left(\frac{3a}{2}\right)^2}{4a} + \dfrac{6mg\left(\frac{5a}{2}\right)^2}{8a} + \dfrac{mga}{2} + \dfrac{mv^2}{2}\) | M1A1 | |
| \(v = \sqrt{\dfrac{15ga}{8}}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Correct use of Hooke's law for either string | B1 | Must include an unknown extension |
| Resolve vertically with two variable tensions and weight | M1 | M0 for setting both extensions as \(e\) |
| Correct equation | A1 | |
| Solve to find either extension | dM1 | |
| Correct extensions found for both strings | A1* | From fully correct working |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Vertical equation of motion with two different variable tensions, weight and \(m\ddot{x}\) | M1 | Allow \(ma\) |
| Equation with at most one error | A1 | Allow \(ma\) for this mark, does not count as error |
| Fully correct equation | A1 | Must now be \(m\ddot{x}\) |
| \(\ddot{x} = -\frac{5g}{2a}x\) \(\therefore\) SHM | A1 | Must have concluding statement |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Use of their \(\omega^2\) | B1ft | |
| Complete method to find speed at \(\frac{7}{2}a\) above \(A\) | M1 | Follow through their \(\omega\). Needs amplitude \(a\) and \(x = \frac{1}{2}a\) |
| Correct equation | A1 | No follow through |
| Correct final answer | A1 | cso |
## Question 5:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dfrac{2mge_1}{2a}$ or $\dfrac{6mg(4a-e_1)}{4a}$ | B1 | |
| $mg + \dfrac{2mge_1}{2a} = \dfrac{6mg(4a-e_1)}{4a}$ | M1A1 | |
| Solve to find either extension | dM1 | |
| $e_1 = 2a$ and $e_2 = 4a - e_1 = 2a$* | A1* | |
**ALT (a):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $mg + \dfrac{2mge_1}{2a} = \dfrac{6mge_2}{4a}$, $e_1 + e_2 = 4a$ | M1A1 | |
| Solve simultaneously | dM1 | |
| $e_1 = 2a$ and $e_2 = 2a$* | A1* | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $mg + \dfrac{2mg(2a-x)}{2a} - \dfrac{6mg(2a+x)}{4a} = m\ddot{x}$ | M1A1A1 | |
| $\ddot{x} = -\dfrac{5g}{2a}x \therefore$ SHM | A1 | |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\omega^2 = \dfrac{5g}{2a}$ | B1ft | |
| $v^2 = \dfrac{5g}{2a}\left(a^2 - \left(\dfrac{a}{2}\right)^2\right)$ | M1A1 | |
| $v = \sqrt{\dfrac{15ga}{8}} = \dfrac{\sqrt{30ga}}{4}$ | A1 cso | |
**ALT (c):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dfrac{2mga^2}{4a}$ or $\dfrac{6mg(3a)^2}{8a}$ or $\dfrac{2mg\left(\frac{3a}{2}\right)^2}{4a}$ or $\dfrac{6mg\left(\frac{5a}{2}\right)^2}{8a}$ | B1 | |
| $\dfrac{2mga^2}{4a} + \dfrac{6mg(3a)^2}{8a} = \dfrac{2mg\left(\frac{3a}{2}\right)^2}{4a} + \dfrac{6mg\left(\frac{5a}{2}\right)^2}{8a} + \dfrac{mga}{2} + \dfrac{mv^2}{2}$ | M1A1 | |
| $v = \sqrt{\dfrac{15ga}{8}}$ | A1 | |
# Question 5 (SHM Question):
## Part (a)
| Working/Answer | Mark | Guidance |
|---|---|---|
| Correct use of Hooke's law for either string | B1 | Must include an unknown extension |
| Resolve vertically with two variable tensions and weight | M1 | M0 for setting both extensions as $e$ |
| Correct equation | A1 | |
| Solve to find either extension | dM1 | |
| Correct extensions found for both strings | A1* | From fully correct working |
## Part (b)
| Working/Answer | Mark | Guidance |
|---|---|---|
| Vertical equation of motion with two different variable tensions, weight and $m\ddot{x}$ | M1 | Allow $ma$ |
| Equation with at most one error | A1 | Allow $ma$ for this mark, does not count as error |
| Fully correct equation | A1 | Must now be $m\ddot{x}$ |
| $\ddot{x} = -\frac{5g}{2a}x$ $\therefore$ SHM | A1 | **Must have concluding statement** |
## Part (c)
| Working/Answer | Mark | Guidance |
|---|---|---|
| Use of their $\omega^2$ | B1ft | |
| Complete method to find speed at $\frac{7}{2}a$ above $A$ | M1 | Follow through their $\omega$. Needs amplitude $a$ and $x = \frac{1}{2}a$ |
| Correct equation | A1 | No follow through |
| Correct final answer | A1 | cso |
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8a687d17-ec7e-463f-84dd-605f5c230db1-16_720_232_251_858}
\captionsetup{labelformat=empty}
\caption{Figure 6}
\end{center}
\end{figure}
The fixed points, $A$ and $B$, are a distance $10 a$ apart, with $B$ vertically above $A$.
One end of a light elastic string, of natural length $2 a$ and modulus of elasticity $2 m g$, is attached to a particle $P$ of mass $m$ and the other end is attached to $A$.
One end of another light elastic string, of natural length $4 a$ and modulus of elasticity $6 m g$, is attached to $P$ and the other end is attached to $B$.
The particle $P$ rests in equilibrium at the point $C$, as shown in Figure 6.
\begin{enumerate}[label=(\alph*)]
\item Show that each string has an extension of $2 a$.\\
(5)
The particle $P$ is now pulled down vertically, so that it is a distance $a$ below $C$ and then released from rest.
\item Show that in the subsequent motion, $P$ performs simple harmonic motion.
\item Find, in terms of $a$ and $g$, the speed of $P$ when it is a distance $\frac { 7 } { 2 } a$ above $A$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2021 Q5 [13]}}