| Exam Board | AQA |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2024 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Difficulty | Challenging +1.2 This is a standard Further Maths mechanics question combining elastic strings, equilibrium, and damped harmonic motion. Part (a) requires routine force balance with Hooke's law. Part (b)(i) is a straightforward 'show that' derivation using F=ma with damping and restoring forces. Part (b)(ii) involves solving a standard second-order linear ODE with given initial conditions—a core Further Maths technique. While multi-step and requiring several concepts, all components are textbook exercises without novel insight required. |
| Spec | 4.10e Second order non-homogeneous: complementary + particular integral4.10g Damped oscillations: model and interpret6.02g Hooke's law: T = k*x or T = lambda*x/l6.02h Elastic PE: 1/2 k x^2 |
| Answer | Marks |
|---|---|
| 18(a) | Forms equilibrium force |
| Answer | Marks | Guidance |
|---|---|---|
| Condone sign errors. | 3.1b | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| extensions. | 3.1b | M1 |
| Obtains 1.39 and 1.61 | 1.1b | A1 |
| Subtotal | 3 | |
| Q | Marking | |
| instructions | AO | Marks |
| Answer | Marks |
|---|---|
| 18(b)(i) | Forms at least one |
| Answer | Marks | Guidance |
|---|---|---|
| FT their e A or e C | 3.1b | B1F |
| Answer | Marks | Guidance |
|---|---|---|
| on the terms. | 3.1b | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| AG | 2.1 | R1 |
| Subtotal | 3 | |
| Q | Marking instructions | AO |
| Answer | Marks | Guidance |
|---|---|---|
| 18(b)(ii) | Obtains solution from their three | |
| term Auxiliary Equation. | 3.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Ae–4t + Be–5t | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| find an equation in two variables | 3.4 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| variables | 3.3 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| x = 3e–4t –2.4e–5t OE | 1.1b | A1 |
| Subtotal | 5 | |
| Q | Marking instructions | AO |
| Answer | Marks | Guidance |
|---|---|---|
| 18(c) | Gives a valid limitation with | |
| reference to the size of the tube. | 3.5b | E1 |
| Answer | Marks |
|---|---|
| Subtotal | 1 |
| Question total | 12 |
| Question paper total | 100 |
Question 18:
--- 18(a) ---
18(a) | Forms equilibrium force
equation with three correct
terms.
Condone sign errors. | 3.1b | M1 | 7e = 3e + 0.5g
A C
3 = e A + e C
e A = 1.39, e C = 1.61
Uses the equation
3 = e A + e C and their force
equation to obtain the
extensions. | 3.1b | M1
Obtains 1.39 and 1.61 | 1.1b | A1
Subtotal | 3
Q | Marking
instructions | AO | Marks | Typical solution
--- 18(b)(i) ---
18(b)(i) | Forms at least one
correct expression in
x for the tension
ie 7(e A + x) or
3(e – x)
C
FT their e A or e C | 3.1b | B1F | 0.5x = 0.5g + 3(e C – x) – 7(e A + x) – 4.5 x
0.5 x = 0.5g + 3(1.61 – x) – 7(1.39 + x) – 4.5 x
0.5 x = 4.9 + 4.83 – 3x – 9.73 – 7x – 4.5 x
x + 9 x + 20x = 0
Forms an equation
of motion with five
terms (with at least
two terms correct).
Accept “a” for x
And “v” for x
Condone sign errors
on the terms. | 3.1b | M1
Completes a
reasoned argument
to obtain
x + 9 x + 20x = 0
OE
AG | 2.1 | R1
Subtotal | 3
Q | Marking instructions | AO | Marks | Typical solution
--- 18(b)(ii) ---
18(b)(ii) | Obtains solution from their three
term Auxiliary Equation. | 3.1a | M1 | 0 = λ2 + 9λ + 20
λ = –4 or λ = –5
x = Ae–4t + Be–5t
x = –4Ae–4t – 5Be–5t
x = 0.6, x = 0, t = 0
0.6 = A + B
0 = –4A – 5B
A = 3 and B = –2.4
x = 3e–4t –2.4e–5t
Obtains
Ae–4t + Be–5t | 1.1b | A1
Uses x = 0.6 when t = 0 to
find an equation in two variables | 3.4 | M1
Sets their correct x = 0 when
t = 0 to find an equation in two
variables | 3.3 | M1
Obtains
x = 3e–4t –2.4e–5t OE | 1.1b | A1
Subtotal | 5
Q | Marking instructions | AO | Marks | Typical solution
--- 18(c) ---
18(c) | Gives a valid limitation with
reference to the size of the tube. | 3.5b | E1 | The resistance force in a thin tube
might not be the same as in a large
bath.
Subtotal | 1
Question total | 12
Question paper total | 100In this question use $g = 9.8$ m s$^{-2}$
Two light elastic strings each have one end attached to a small ball $B$ of mass 0.5 kg
The other ends of the strings are attached to the fixed points $A$ and $C$, which are 8 metres apart with $A$ vertically above $C$
The whole system is in a thin tube of oil, as shown in the diagram below.
\includegraphics{figure_18}
The string connecting $A$ and $B$ has natural length 2 metres, and the tension in this string is $7e$ newtons when the extension is $e$ metres.
The string connecting $B$ and $C$ has natural length 3 metres, and the tension in this string is $3e$ newtons when the extension is $e$ metres.
\begin{enumerate}[label=(\alph*)]
\item Find the extension of each string when the system is in equilibrium. [3 marks]
\item It is known that in a large bath of oil, the oil causes a resistive force of magnitude $4.5v$ newtons to act on the ball, where $v$ m s$^{-1}$ is the speed of the ball.
Use this model to answer part (b)(i) and part (b)(ii).
\begin{enumerate}[label=(\roman*)]
\item The ball is pulled a distance of 0.6 metres downwards from its equilibrium position towards $C$, and released from rest.
Show that during the subsequent motion the particle satisfies the differential equation
$$\frac{d^2x}{dt^2} + 9\frac{dx}{dt} + 20x = 0$$
where $x$ metres is the displacement of the particle below the equilibrium position at time $t$ seconds after the particle is released. [3 marks]
\item Find $x$ in terms of $t$ [5 marks]
\end{enumerate}
\item State one limitation of the model used in part (b) [1 mark]
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 1 2024 Q18 [12]}}