| Exam Board | Edexcel |
|---|---|
| Module | FM2 (Further Mechanics 2) |
| Year | 2019 |
| Session | June |
| 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 Further Mechanics 2 SHM question requiring proof of SHM from Hooke's law, finding amplitude and period, then using standard SHM formulas. While it involves multiple parts and careful setup with two string sections, the techniques are routine for FM2 students: establish equilibrium, apply F=-kx, then apply standard SHM results. Part (c) requires inverse trig but follows a standard template. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x6.02g Hooke's law: T = k*x or T = lambda*x/l |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use of Hooke's law seen | B1 | Use of Hooke's law seen at least once (with natural length \(\leq 1.5\)) |
| Let \(x\) be displacement from equilibrium, then \(T_B - T_A = -0.2\ddot{x}\) | M1 | Form equation of motion involving difference of 2 tensions; must be dimensionally correct; need all terms; condone incorrect lengths; condone sign errors |
| \(\frac{15(2+x)}{1} - \frac{15(1-x)}{0.5} = -0.2\ddot{x}\) | A1 | Unsimplified equation with at most one error |
| A1 | Correct unsimplified equation | |
| \(\Rightarrow \ddot{x} = -225x = -15^2x\), which is of the form \(\ddot{x}=-\omega^2 x\), so SHM * | A1* | All correct and justification of SHM – comment required |
| (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(a = 0.5\), \(\omega = 15\) or their \(\omega\) | B1ft | Correct values seen or implied; \(a\) must be correct but follow their \(\omega\) from (a) |
| Max speed \(= a\omega = 7.5\) (m s\(^{-1}\)) | B1ft | Follow their \(a\), \(\omega\) |
| Max acceleration \(= a\omega^2 = 112.5\) (m s\(^{-2}\)) | B1ft | Follow their \(a\), \(\omega\) |
| (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = 0.5\cos 15t \Rightarrow \dot{x} = -7.5\sin 15t\) | B1ft | Use correct equation for speed \((-a\omega\sin\omega t)\); follow their \(a\) and \(\omega\); allow sin or cos form; can also use correct expression for \(x\) combined with \(v^2 = \omega^2(a^2-x^2)\) |
| \( | \dot{x} | = 7.5\sin 15t = 5 \Rightarrow \sin 15t = \frac{2}{3} \Rightarrow t = \ldots\) |
| \(t = 0.0486\ldots\) (s) | A1 | Seen or implied |
| Complete strategy to find the required time | M1 | Complete strategy – e.g. find value for \(t\) when \(v=5\) and use symmetry and periodic time |
| Speed \(> 5\) for \(\frac{2\pi}{15} - 4t = 0.22\) (s) | A1 | 2 s.f. or better (0.22428...) |
| (5) | (13 marks) |
## Question 6(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of Hooke's law seen | B1 | Use of Hooke's law seen at least once (with natural length $\leq 1.5$) |
| Let $x$ be displacement from equilibrium, then $T_B - T_A = -0.2\ddot{x}$ | M1 | Form equation of motion involving difference of 2 tensions; must be dimensionally correct; need all terms; condone incorrect lengths; condone sign errors |
| $\frac{15(2+x)}{1} - \frac{15(1-x)}{0.5} = -0.2\ddot{x}$ | A1 | Unsimplified equation with at most one error |
| | A1 | Correct unsimplified equation |
| $\Rightarrow \ddot{x} = -225x = -15^2x$, which is of the form $\ddot{x}=-\omega^2 x$, so SHM * | A1* | All correct and justification of SHM – comment required |
| **(5)** | | |
## Question 6(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $a = 0.5$, $\omega = 15$ or their $\omega$ | B1ft | Correct values seen or implied; $a$ must be correct but follow their $\omega$ from (a) |
| Max speed $= a\omega = 7.5$ (m s$^{-1}$) | B1ft | Follow their $a$, $\omega$ |
| Max acceleration $= a\omega^2 = 112.5$ (m s$^{-2}$) | B1ft | Follow their $a$, $\omega$ |
| **(3)** | | |
## Question 6(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = 0.5\cos 15t \Rightarrow \dot{x} = -7.5\sin 15t$ | B1ft | Use correct equation for speed $(-a\omega\sin\omega t)$; follow their $a$ and $\omega$; allow sin or cos form; can also use correct expression for $x$ combined with $v^2 = \omega^2(a^2-x^2)$ |
| $|\dot{x}| = 7.5\sin 15t = 5 \Rightarrow \sin 15t = \frac{2}{3} \Rightarrow t = \ldots$ | M1 | Find a relevant time e.g. time taken for 0 m s$^{-1}$ to 5 m s$^{-1}$ |
| $t = 0.0486\ldots$ (s) | A1 | Seen or implied |
| Complete strategy to find the required time | M1 | Complete strategy – e.g. find value for $t$ when $v=5$ and use symmetry and periodic time |
| Speed $> 5$ for $\frac{2\pi}{15} - 4t = 0.22$ (s) | A1 | 2 s.f. or better (0.22428...) |
| **(5)** | **(13 marks)** | |\begin{enumerate}
\item The points $A$ and $B$ lie on a smooth horizontal surface with $A B = 4.5 \mathrm {~m}$.
\end{enumerate}
A light elastic string has natural length 1.5 m and modulus of elasticity 15 N . One end of the string is attached to $A$ and the other end of the string is attached to $B$. A particle, $P$, of mass 0.2 kg , is attached to the stretched string so that $A P B$ is a straight line and $A P = 1.5 \mathrm {~m}$. The particle rests in equilibrium on the surface.
The particle is now moved directly towards $A$ and is held on the surface so $A P B$ is a straight line with $A P = 1 \mathrm {~m}$.
The particle is released from rest.\\
(a) Prove that $P$ moves with simple harmonic motion.\\
(b) Find\\
(i) the maximum speed of $P$ during the motion,\\
(ii) the maximum acceleration of $P$ during the motion.\\
(c) Find the total time, in each complete oscillation of $P$, for which the speed of $P$ is greater than $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\hfill \mbox{\textit{Edexcel FM2 2019 Q6 [13]}}