Question 13 - A-Level Maths

OCR MEI Further Mechanics Major 2023 June — Question 13 15 marks

Exam Board	OCR MEI
Module	Further Mechanics Major (Further Mechanics Major)
Year	2023
Session	June
Marks	15
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	Challenging +1.2 This is a structured Further Maths mechanics question with clear scaffolding through parts (a)-(f). While it involves SHM with a non-constant forcing term and requires setting up equations of motion in an accelerating reference frame, each part provides significant guidance. The verification in part (e) is routine differentiation, and the final comment requires only sign analysis of the given solution. More challenging than standard A-level due to the Further Maths content and multi-step nature, but the heavy scaffolding prevents it from being truly difficult.
Spec	4.10f Simple harmonic motion: x'' = -omega^2 x 4.10g Damped oscillations: model and interpret 6.02e Calculate KE and PE: using formulae 6.02h Elastic PE: 1/2 k x^2

A particle P of mass $m$ is fixed to one end of a light spring of natural length $a$ and modulus of elasticity $man^2$, where $n > 0$. The other end of the spring is attached to the ceiling of a lift. The lift is at rest and P is hanging vertically in equilibrium.

Find, in terms of $g$ and $n$, the extension in the spring. [3]

At time $t = 0$ the lift begins to accelerate upwards from rest. At time $t$, the upward displacement of the lift from its initial position is $y$ and the extension of the spring is $x$.

Express, in terms of $g$, $n$, $x$ and $y$, the upward displacement of P from its initial position at time $t$. [2]
Given that $\ddot{y} = kt$, where $k$ is a positive constant, express the upward acceleration of P in terms of $\ddot{x}$, $k$ and $t$. [1]
Show that $x$ satisfies the differential equation $$\ddot{x} + n^2 x = kt + g.$$ [3]
Verify that $x = \frac{1}{n^2}(knt + gn - k \sin(nt))$. [4]
By considering $\ddot{x}$ comment on the motion of P relative to the ceiling of the lift for all times after the lift begins to move. [2]

Show mark scheme Show mark scheme source

Question 13:

Answer	Marks	Guidance
13	(a)	T = m g
0	M1	1.1

m a n 2 e

T = 0

Answer	Marks	Guidance
0 a	M1	3.3

0 spring from its natural length

m a n 2 e

0 = m g  e = g n − 2

Answer	Marks	Guidance
a 0	A1	1.1

[3]

Answer	Marks	Guidance
(b)	y + a + e = a + x + z
0	M1	3.1b

upward displacement of P from its initial position (at

time t) – must include terms in y, x and e (allow sign

0 errors only)

Answer	Marks	Guidance
( z = ) y + g n − 2 − x	A1	1.1

[2]

Answer	Marks	Guidance
(c)	( z = ) k t − x	B1

[1]

Answer	Marks	Guidance
(d)	T − m g = m z	M1*

allow sign errors (allow for T − m g = m a ) – implied

by a correct equation

m a n 2 x

− m g = m ( k t − x )

a

Answer	Marks	Guidance
or mx=mkt+mg−mn2x	M1dep*	3.4

equal to their expression from part (c) – if (c) blank

then must be correct

Answer	Marks	Guidance
n 2 x − g = k t − x  x + n 2 x = k t + g	A1	2.2a

errors then A0

[3]

Answer	Marks
(e)	1

x ( k n t g n k s i n ( n t ) )

Answer	Marks
n 3	For reference: x + n 2 x = k t + g and

1 x = ( k n t + g n − k s i n ( n t ) )

n 3

1 x= (kn−kncos(nt))

Answer	Marks	Guidance
n3	B1	1.1

1 x = ( k n 2 s i n ( n t ) )

Answer	Marks	Guidance
n 3	B1ft	1.1

1  1 

k ( n 2 s i n n ( t ) ) + n 2 k n ( t + g n − k s i n ( n t )

n 3 n 3

k s i n ( n t ) k t n + g n − k s i n ( n t )

= + = k t + g

Answer	Marks	Guidance
n n	B1	1.1

least one stage of working before obtaining kt + g

1 When t = 0, x ( g n ) g n 2

n 3

1 and x = ( k n − k n c o s ( 0 ) ) = 0

Answer	Marks	Guidance
n 3	B1	3.4

[4]

Answer	Marks
CF: A c o s n t + B s i n n t	B1*

kt g

GS: x=Acosnt+Bsinnt+ +

Answer	Marks
n2 n2	kt g

GS: x=Acosnt+Bsinnt+ +

Answer	Marks	Guidance
n2 n2	B1*	B1*

g k

x t x , x 0 ,      = +  = =  = =

n 2 n 2

g

Initial conditions: t = 0 , x = and x=0

Answer	Marks
n 2	B1dep*

k k kt g

A=0,B=− x=− sinnt+ +

Answer	Marks
n3 n3 n2 n2	B1

We can therefore infer that P does not move closer to the

Answer	Marks	Guidance
ceiling of the lift in the subsequent motion	B1dep*	2.2b

P is never nearer the ceiling of the lift as it was when

t = 0, or ‘is always moving away from the ceiling’

[2]

PMT

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Question 13:
13 | (a) | T = m g
0 | M1 | 1.1 | Resolving vertically for P
m a n 2 e
T = 0
0 a | M1 | 3.3 | Apply Hooke’s law – where e is the extension of the
0
spring from its natural length
m a n 2 e
0 = m g  e = g n − 2
a 0 | A1 | 1.1
[3]
(b) | y + a + e = a + x + z
0 | M1 | 3.1b | Attempt an equation containing z where z is the
upward displacement of P from its initial position (at
time t) – must include terms in y, x and e (allow sign
0
errors only)
( z = ) y + g n − 2 − x | A1 | 1.1
[2]
(c) | ( z = ) k t − x | B1 | 1.1
[1]
(d) | T − m g = m z | M1* | 2.1 | Apply N2L vertically – correct number of terms but
allow sign errors (allow for T − m g = m a ) – implied
by a correct equation
m a n 2 x
− m g = m ( k t − x )
a
or mx=mkt+mg−mn2x | M1dep* | 3.4 | Use Hooke’s law with extension = x and acceleration
equal to their expression from part (c) – if (c) blank
then must be correct
n 2 x − g = k t − x  x + n 2 x = k t + g | A1 | 2.2a | AG – so sufficient working must be shown – any
errors then A0
[3]
(e) | 1
x ( k n t g n k s i n ( n t ) )
n 3 | For reference: x + n 2 x = k t + g and
1
x = ( k n t + g n − k s i n ( n t ) )
n 3
1
x= (kn−kncos(nt))
n3 | B1 | 1.1 | Correct derivative
1
x = ( k n 2 s i n ( n t ) )
n 3 | B1ft | 1.1 | Follow through their first derivative
1  1 
k ( n 2 s i n n ( t ) ) + n 2 k n ( t + g n − k s i n ( n t )
n 3 n 3
k s i n ( n t ) k t n + g n − k s i n ( n t )
= + = k t + g
n n | B1 | 1.1 | Correctly confirm given differential equation - at
least one stage of working before obtaining kt + g
1
When t = 0, x ( g n ) g n 2
n 3
1
and x = ( k n − k n c o s ( 0 ) ) = 0
n 3 | B1 | 3.4 | Confirm initial conditions
[4]
CF: A c o s n t + B s i n n t | B1*
kt g
GS: x=Acosnt+Bsinnt+ +
n2 n2 | kt g
GS: x=Acosnt+Bsinnt+ +
n2 n2 | B1* | B1* | From a trial solution of the form
g k
x t x , x 0 ,      = +  = =  = =
n 2 n 2
g
Initial conditions: t = 0 , x = and x=0
n 2 | B1dep*
k k kt g
A=0,B=− x=− sinnt+ +
n3 n3 n2 n2 | B1
We can therefore infer that P does not move closer to the
ceiling of the lift in the subsequent motion | B1dep* | 2.2b | or equivalent comment e.g. in the subsequent motion
P is never nearer the ceiling of the lift as it was when
t = 0, or ‘is always moving away from the ceiling’
[2]
PMT
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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.
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Alternatively, you can email us on
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ocr.org.uk
Twitter/ocrexams
/ocrexams
/company/ocr
/ocrexams
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For staff training purposes and as part of our quality assurance programme your call may be recorded or monitored. © OCR
2023 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

A particle P of mass $m$ is fixed to one end of a light spring of natural length $a$ and modulus of elasticity $man^2$, where $n > 0$. The other end of the spring is attached to the ceiling of a lift. The lift is at rest and P is hanging vertically in equilibrium.

\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $g$ and $n$, the extension in the spring. [3]
\end{enumerate}

At time $t = 0$ the lift begins to accelerate upwards from rest. At time $t$, the upward displacement of the lift from its initial position is $y$ and the extension of the spring is $x$.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Express, in terms of $g$, $n$, $x$ and $y$, the upward displacement of P from its initial position at time $t$. [2]
\item Given that $\ddot{y} = kt$, where $k$ is a positive constant, express the upward acceleration of P in terms of $\ddot{x}$, $k$ and $t$. [1]
\item Show that $x$ satisfies the differential equation
$$\ddot{x} + n^2 x = kt + g.$$ [3]
\item Verify that $x = \frac{1}{n^2}(knt + gn - k \sin(nt))$. [4]
\item By considering $\ddot{x}$ comment on the motion of P relative to the ceiling of the lift for all times after the lift begins to move. [2]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Mechanics Major 2023 Q13 [15]}}

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Q1 4 Q2 4 Q3 5 Q4 6 Q5 7 Q6 9 Q7 9 Q8 8 Q9 12 Q10 16 Q11 12 Q12 13 Q13 15