| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics B AS (Further Mechanics B AS) |
| Year | 2022 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Prove SHM and find period: given force or equation of motion directly |
| Difficulty | Standard +0.3 This is a standard SHM problem requiring application of Hooke's law and Newton's second law to derive a differential equation, then transforming to standard SHM form. The steps are routine for Further Maths students: find tension using extension, apply F=ma, identify equilibrium position, and substitute. The algebra is straightforward and the method is well-practiced in this topic. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x4.10g Damped oscillations: model and interpret6.02g Hooke's law: T = k*x or T = lambda*x/l |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | (a) | 𝑘𝑚𝑔(𝑦−𝑎) |
| Answer | Marks |
|---|---|
| 𝑎 d𝑡2 | M1 |
| A1 | 3.3 |
| 1.1 | M1 for expression for tension with |
| Answer | Marks |
|---|---|
| Correct Newton 2 law | The tension term has to |
| Answer | Marks | Guidance |
|---|---|---|
| d𝑡2 𝑎 | A1 | 1.1 |
| Answer | Marks |
|---|---|
| (b) | d2𝑦 𝑘𝑔 𝑙 |
| Answer | Marks | Guidance |
|---|---|---|
| d𝑡2 𝑎 𝑘 | M1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 𝑘 | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 𝑎 | B1 | 1.1 |
| ALT | M1 | Find the equilibrium point of the |
| Answer | Marks |
|---|---|
| A1 | Finding relationship between x and y |
| and rearranging correctly | x is the distance of the point |
| Answer | Marks |
|---|---|
| B1 | Correct answer enough for B1 here |
Question 2:
2 | (a) | 𝑘𝑚𝑔(𝑦−𝑎)
𝑇 =
𝑎
𝑘𝑚𝑔(𝑦−𝑎) d2𝑦
𝑚𝑔− = 𝑚
𝑎 d𝑡2 | M1
A1 | 3.3
1.1 | M1 for expression for tension with
attempt at extension (rather than y)
Correct Newton 2 law | The tension term has to
include the extension (either
as y – a or another letter
identified as extension)
𝑘𝑔(𝑦−𝑎) d2𝑦
𝑔− =
𝑎 d𝑡2
𝑘𝑔𝑦 d2𝑦
𝑔− +𝑘𝑔 =
𝑎 d𝑡2
d2𝑦 𝑘𝑔
= (𝑘 +1)𝑔− 𝑦
d𝑡2 𝑎 | A1 | 1.1 | AG | Must have (at least) one line
of working in between
correct Newton 2 and the
given answer, which shows
the decomposition of the
tension term.
[3]
(b) | d2𝑦 𝑘𝑔 𝑙
= − (𝑦− (𝑘 +1)).
d𝑡2 𝑎 𝑘 | M1 | 1.1 | Oe
𝑎
𝑥 = 𝑦− (𝑘+1)
𝑘 | A1 | 1.1
𝑘𝑔
𝜔 = √( )
𝑎 | B1 | 1.1
ALT | M1 | Find the equilibrium point of the
motion
A1 | Finding relationship between x and y
and rearranging correctly | x is the distance of the point
below the equilibrium point
Find the equilibrium point of the
motion
Finding relationship between x and y
and rearranging correctly
x is the distance of the point
below the equilibrium point
B1 | Correct answer enough for B1 here
[3]2 A light elastic string has natural length $a$ and modulus of elasticity $k m g$, where $k > 2$. One end of the string is attached to a fixed point O . A particle P of mass $m$ is attached to the other end of the string. P is held at rest a distance $\frac { 3 } { 2 } a$ vertically below O .
At time $t$ after P is released, its vertical distance below O is $y$.
\begin{enumerate}[label=(\alph*)]
\item Show that, while the string is in tension, the equation of motion of P is given by the differential equation $\frac { d ^ { 2 } y } { d t ^ { 2 } } = ( k + 1 ) g - \frac { k g } { a } y$.
A student transforms the differential equation in part (a) into the standard SHM equation $\frac { d ^ { 2 } x } { d t ^ { 2 } } = - \omega ^ { 2 } x$.
\item - Find an expression for $x$ in terms of $y , k$ and $a$.
\begin{itemize}
\item Find an expression for $\omega$ in terms of $k , a$ and $g$.
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics B AS 2022 Q2 [6]}}