Question 15 - A-Level Maths

OCR MEI Further Pure Core 2022 June — Question 15 23 marks

Exam Board	OCR MEI
Module	Further Pure Core (Further Pure Core)
Year	2022
Session	June
Marks	23
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Second order differential equations
Type	Modeling context with interpretation
Difficulty	Challenging +1.2 This is a structured, multi-part question on second-order differential equations with physical context. While it covers SHM, damped oscillations, and forced oscillations comprehensively, each part provides significant scaffolding. The 'show that' parts guide students to standard forms, the classification tasks are routine, and solving the equations uses standard auxiliary equation methods. The final verification requires recognizing transient decay, but this is a standard observation in forced oscillations. More challenging than typical A-level due to Further Maths content and length, but less demanding than questions requiring novel insight or complex manipulation.
Spec	4.10e Second order non-homogeneous: complementary + particular integral 4.10f Simple harmonic motion: x'' = -omega^2 x 4.10g Damped oscillations: model and interpret

15 In an oscillating system, a particle of mass \(m \mathrm {~kg}\) moves in a horizontal line. Its displacement from its equilibrium position O at time \(t\) seconds is \(x\) metres, its velocity is \(v \mathrm {~ms} ^ { - 1 }\), and it is acted on by a force \(2 m x\) newtons acting towards O as shown in the diagram. \includegraphics[max width=\textwidth, alt={}, center]{b57a2590-84e8-4998-9633-902db861f23a-6_212_914_408_242} Initially, the particle is projected away from O with speed \(1 \mathrm {~ms} ^ { - 1 }\) from a point 2 m from O in the positive direction.

1. Show that the motion is modelled by the differential equation \(\frac { d ^ { 2 } x } { d t ^ { 2 } } + 2 x = 0\).
2. State the type of motion.
3. Write down the period of the motion.
4. Find \(x\) in terms of \(t\).
5. Find the amplitude of the motion.
The motion is now damped by a force \(2 m v\) newtons.
1. Show that \(\frac { d ^ { 2 } x } { d t ^ { 2 } } + 2 \frac { d x } { d t } + 2 x = 0\).
2. State, giving a reason, whether the system is under-damped, critically damped or over-damped.
3. Determine the general solution of this differential equation.
Finally, a variable force \(2 m \cos 2 t\) newtons is added, so that the motion is now modelled by the differential equation \(\frac { d ^ { 2 } x } { d t ^ { 2 } } + 2 \frac { d x } { d t } + 2 x = 2 \cos 2 t\).
1. Find \(x\) in terms of \(t\). In the long term, the particle is seen to perform simple harmonic motion with a period of just over 3 seconds.
2. Verify that this behaviour is consistent with the answer to part (c)(i).

Show mark scheme Show mark scheme source

Question 15:

Answer	Marks	Guidance
15	(a)	(i)

𝑚 = −2𝑚𝑥 ⇒ +2𝑥 = 0

Answer	Marks	Guidance
d𝑡2 d𝑡2	B1
[1]	3.1b	AG
15	(a)	(ii)
[1]	3.3
15	(a)	(iii)

= √2𝜋 (s)

Answer	Marks	Guidance
√2	B1
[1]	1.2	Accept anything wrt 4.4
15	(a)	(iv)

𝑡 = 0, 𝑥 = 2 ⇒ 𝐴 = 2

d𝑥

= −√2𝐴sin√2𝑡+√2𝐵cos√2𝑡

d𝑡

d𝑥 1 √2

𝑡 = 0, = 1 ⇒ 𝐵 = =

d𝑡 √2 2

√2

so 𝑥 = 2cos√2𝑡+ sin√2𝑡

Answer	Marks
2	B1

B1

M1

A1

Answer	Marks
[4]	1.1

3.3

1.1

Answer	Marks
3.3	Must come from correct GS

Must be differentiating a function in terms of cos and sin

Answer	Marks	Guidance
15	(a)	(v)

Amplitude = √22+( √2)

2 3√2

= (m)

Answer	Marks
2	M1

A1

Answer	Marks	Guidance
[2]	3.4
1.1	Accept anything wrt 2.1
15	(b)	(i)

𝑚 = −2𝑚𝑥−2𝑚

d𝑡2 d𝑡

d2𝑥 d𝑥

⇒ +2 +2𝑥 = 0

Answer	Marks	Guidance
d𝑡2 d𝑡	B1
[1]	3.1b	AG
15	(b)	(ii)
[1]	3.5b	Stating complex roots of auxillary equation is acceptable

(−1±𝑖)

Answer	Marks	Guidance
15	(b)	(iii)

−2±√−4

𝜆 = = −1±𝑖

2

Answer	Marks
General solution: 𝑥 = 𝑒−𝑡(𝐴cos𝑡+𝐵sin𝑡)	M1

A1

Answer	Marks
[3]	1.2

1.1

Answer	Marks	Guidance
1.1	soi
15	(c)	(i)

−4𝐶+4𝐷+2𝐶 = 2, −4𝐷−4𝐶+2𝐷 = 0

⇒ 𝐶 = −0.2, 𝐷 = 0.4

𝑥 = 𝑒−𝑡(𝐴cos𝑡+𝐵sin𝑡)−0.2cos2𝑡+0.4sin2𝑡

𝑡 = 0, 𝑥 = 2 ⇒ 𝐴 = 2.2

d𝑥

= −𝑒−𝑡(𝐴cos𝑡+𝐵sin𝑡)+𝑒−𝑡(−𝐴sin𝑡

d𝑡

+𝐵cos𝑡)

+0.4sin2𝑡+0.8cos2𝑡

d𝑥

𝑡 = 0, = 1 ⇒ 1 = −𝐴+𝐵+0.8

d𝑡

⇒ 𝐵 = 2.4

𝑥 = e−𝑡(2.2cos𝑡+2.4sin𝑡)−0.2cos2𝑡

Answer	Marks
+0.4sin2𝑡	M1

M1

A1

B1

M1*

M1de

p*

A1cao

Answer	Marks
[7]	2.1

3.3 2.2a

3.3

2.1

3.3

Answer	Marks
2.2a	Has to come from a correct CF

Must include use of product rule from x = CF + PI

For substitution to lead to an equation in A and B

Answer	Marks	Guidance
15	(c)	(ii)

2𝜋

[this is SHM] with period = 𝜋 ≈ 3.14 𝑠

Answer	Marks
2	M1

A1

Answer	Marks
[2]	3.5a
3.5a	Dependent on a function containing 𝑒−𝑘𝑡and cos𝑝𝑡 and sin𝑝𝑡

PMT

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Question 15:
15 | (a) | (i) | 𝑑2𝑥 d2𝑥
𝑚 = −2𝑚𝑥 ⇒ +2𝑥 = 0
d𝑡2 d𝑡2 | B1
[1] | 3.1b | AG
15 | (a) | (ii) | Simple harmonic motion | B1
[1] | 3.3
15 | (a) | (iii) | 2𝜋
= √2𝜋 (s)
√2 | B1
[1] | 1.2 | Accept anything wrt 4.4
15 | (a) | (iv) | General solution: 𝑥 = 𝐴cos√2𝑡+𝐵sin√2𝑡
𝑡 = 0, 𝑥 = 2 ⇒ 𝐴 = 2
d𝑥
= −√2𝐴sin√2𝑡+√2𝐵cos√2𝑡
d𝑡
d𝑥 1 √2
𝑡 = 0, = 1 ⇒ 𝐵 = =
d𝑡 √2 2
√2
so 𝑥 = 2cos√2𝑡+ sin√2𝑡
2 | B1
B1
M1
A1
[4] | 1.1
3.3
1.1
3.3 | Must come from correct GS
Must be differentiating a function in terms of cos and sin
15 | (a) | (v) | 1 2
Amplitude = √22+( √2)
2
3√2
= (m)
2 | M1
A1
[2] | 3.4
1.1 | Accept anything wrt 2.1
15 | (b) | (i) | d2𝑥 d𝑥
𝑚 = −2𝑚𝑥−2𝑚
d𝑡2 d𝑡
d2𝑥 d𝑥
⇒ +2 +2𝑥 = 0
d𝑡2 d𝑡 | B1
[1] | 3.1b | AG
15 | (b) | (ii) | Underdamped as 22−4×1×2 < 0 | B1
[1] | 3.5b | Stating complex roots of auxillary equation is acceptable
(−1±𝑖)
15 | (b) | (iii) | Auxiliary equation: 𝜆2+2𝜆+2 = 0
−2±√−4
𝜆 = = −1±𝑖
2
General solution: 𝑥 = 𝑒−𝑡(𝐴cos𝑡+𝐵sin𝑡) | M1
A1
A1
[3] | 1.2
1.1
1.1 | soi
15 | (c) | (i) | Particular integral: 𝑥 = 𝐶𝑐𝑜𝑠2𝑡+𝐷𝑠𝑖𝑛2𝑡
−4𝐶+4𝐷+2𝐶 = 2, −4𝐷−4𝐶+2𝐷 = 0
⇒ 𝐶 = −0.2, 𝐷 = 0.4
𝑥 = 𝑒−𝑡(𝐴cos𝑡+𝐵sin𝑡)−0.2cos2𝑡+0.4sin2𝑡
𝑡 = 0, 𝑥 = 2 ⇒ 𝐴 = 2.2
d𝑥
= −𝑒−𝑡(𝐴cos𝑡+𝐵sin𝑡)+𝑒−𝑡(−𝐴sin𝑡
d𝑡
+𝐵cos𝑡)
+0.4sin2𝑡+0.8cos2𝑡
d𝑥
𝑡 = 0, = 1 ⇒ 1 = −𝐴+𝐵+0.8
d𝑡
⇒ 𝐵 = 2.4
𝑥 = e−𝑡(2.2cos𝑡+2.4sin𝑡)−0.2cos2𝑡
+0.4sin2𝑡 | M1
M1
A1
B1
M1*
M1de
p*
A1cao
[7] | 2.1
3.3
2.2a
3.3
2.1
3.3
2.2a | Has to come from a correct CF
Must include use of product rule from x = CF + PI
For substitution to lead to an equation in A and B
15 | (c) | (ii) | As 𝑡 → ∞, 𝑥 → −0.2cos2𝑡+0.4sin2𝑡
2𝜋
[this is SHM] with period = 𝜋 ≈ 3.14 𝑠
2 | M1
A1
[2] | 3.5a
3.5a | Dependent on a function containing 𝑒−𝑘𝑡and cos𝑝𝑡 and sin𝑝𝑡
PMT
Need to get in touch?
If you ever have any questions about OCR qualifications or services (including administration, logistics and teaching) please feel free to get in
touch with our customer support centre.
Call us on
01223 553998
Alternatively, you can email us on
support@ocr.org.uk
For more information visit
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ocr.org.uk
Twitter/ocrexams
/ocrexams
/company/ocr
/ocrexams
OCR is part of Cambridge University Press & Assessment, a department of the University of Cambridge.
For staff training purposes and as part of our quality assurance programme your call may be recorded or monitored. © OCR
2022 Oxford Cambridge and RSA Examinations is a Company Limited by Guarantee. Registered in England. Registered office
The Triangle Building, Shaftesbury Road, Cambridge, CB2 8EA.
Registered company number 3484466. OCR is an exempt charity.
OCR operates academic and vocational qualifications regulated by Ofqual, Qualifications Wales and CCEA as listed in their
qualifications registers including A Levels, GCSEs, Cambridge Technicals and Cambridge Nationals.
OCR provides resources to help you deliver our qualifications. These resources do not represent any particular teaching method
we expect you to use. We update our resources regularly and aim to make sure content is accurate but please check the OCR
website so that you have the most up-to-date version. OCR cannot be held responsible for any errors or omissions in these
resources.
Though we make every effort to check our resources, there may be contradictions between published support and the
specification, so it is important that you always use information in the latest specification. We indicate any specification changes
within the document itself, change the version number and provide a summary of the changes. If you do notice a discrepancy
between the specification and a resource, please contact us.
Whether you already offer OCR qualifications, are new to OCR or are thinking about switching, you can request more
information using our Expression of Interest form.
Please get in touch if you want to discuss the accessibility of resources we offer to support you in delivering our qualifications.

Show LaTeX source

15 In an oscillating system, a particle of mass $m \mathrm {~kg}$ moves in a horizontal line. Its displacement from its equilibrium position O at time $t$ seconds is $x$ metres, its velocity is $v \mathrm {~ms} ^ { - 1 }$, and it is acted on by a force $2 m x$ newtons acting towards O as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{b57a2590-84e8-4998-9633-902db861f23a-6_212_914_408_242}

Initially, the particle is projected away from O with speed $1 \mathrm {~ms} ^ { - 1 }$ from a point 2 m from O in the positive direction.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Show that the motion is modelled by the differential equation $\frac { d ^ { 2 } x } { d t ^ { 2 } } + 2 x = 0$.
\item State the type of motion.
\item Write down the period of the motion.
\item Find $x$ in terms of $t$.
\item Find the amplitude of the motion.
\end{enumerate}\item The motion is now damped by a force $2 m v$ newtons.
\begin{enumerate}[label=(\roman*)]
\item Show that $\frac { d ^ { 2 } x } { d t ^ { 2 } } + 2 \frac { d x } { d t } + 2 x = 0$.
\item State, giving a reason, whether the system is under-damped, critically damped or over-damped.
\item Determine the general solution of this differential equation.
\end{enumerate}\item Finally, a variable force $2 m \cos 2 t$ newtons is added, so that the motion is now modelled by the differential equation\\
$\frac { d ^ { 2 } x } { d t ^ { 2 } } + 2 \frac { d x } { d t } + 2 x = 2 \cos 2 t$.
\begin{enumerate}[label=(\roman*)]
\item Find $x$ in terms of $t$.

In the long term, the particle is seen to perform simple harmonic motion with a period of just over 3 seconds.
\item Verify that this behaviour is consistent with the answer to part (c)(i).
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Pure Core 2022 Q15 [23]}}

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Q1 7 Q2 5 Q3 6 Q4 7 Q5 7 Q6 5 Q7 9 Q8 11 Q9 12 Q10 10 Q11 8 Q12 9 Q13 17 Q14 8 Q15 23