OCR MEI Further Pure Core Specimen — Question 16 18 marks

Exam BoardOCR MEI
ModuleFurther Pure Core (Further Pure Core)
SessionSpecimen
Marks18
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSimple Harmonic Motion
TypeProve motion is SHM from equation
DifficultyChallenging +1.2 This is a substantial multi-part question on damped harmonic motion requiring SHM differential equations, solving second-order ODEs with complex roots, and applying boundary conditions. While it involves several techniques and 18 marks total, each individual part follows standard Further Maths procedures: writing SHM equations from period, solving characteristic equations, and using exponential decay properties. The conceptual demand is moderate—understanding damping and amplitude reduction—but the mathematical steps are routine for Further Pure students.
Spec4.10f Simple harmonic motion: x'' = -omega^2 x4.10g Damped oscillations: model and interpret

A small object is attached to a spring and performs oscillations in a vertical line. The displacement of the object at time \(t\) seconds is denoted by \(x\) cm. Preliminary observations suggest that the object performs simple harmonic motion (SHM) with a period of 2 seconds about the point at which \(x = 0\).
    1. Write down a differential equation to model this motion. [3]
    2. Give the general solution of the differential equation in part (i) (A). [1]
Subsequent observations indicate that the object's motion would be better modelled by the differential equation $$\frac{d^2x}{dt^2} + 2k \frac{dx}{dt} + (k^2 + 9)x = 0 \qquad (*)$$ where \(k\) is a positive constant.
    1. Obtain the general solution of (*). [3]
    2. State two ways in which the motion given by this model differs from that in part (i). [2]
The amplitude of the object's motion is observed to reduce with a scale factor of 0.98 from one oscillation to the next.
  1. Find the value of \(k\). [3]
At the start of the object's motion, \(x = 0\) and the velocity is 12 cm s\(^{-1}\) in the positive \(x\) direction.
  1. Find an equation for \(x\) as a function of \(t\). [4]
  2. Without doing any further calculations, explain why, according to this model, the greatest distance of the object from its starting point in the subsequent motion will be slightly less than 4 cm. [2]

Question 16:
AnswerMarks Guidance
16(i) (A)
T (cid:32) with T (cid:32)2
(cid:90)
(cid:90)(cid:32)(cid:83)
d2x
(cid:32)(cid:16)(cid:83)2x
AnswerMarks
dt2M1
A1
B1
AnswerMarks
[3]3.1b
1.1
3.3
AnswerMarks Guidance
16(i) (B)
[1]1.1 n
or e.g. x(cid:32) Asin((cid:83)t(cid:14)(cid:72))
AnswerMarks Guidance
16(ii) (A)
Roots (cid:79)(cid:32)(cid:16)k(cid:114)3i
AnswerMarks
Hence x(cid:32)e(cid:16)kt(Acos3t(cid:14)Bsin3t)M1
A1
A1
AnswerMarks
[3]1.1a
m
1.1
AnswerMarks Guidance
2.2ae
16(ii) (B)
… and period is a bit more than 2sc
E1
E1
AnswerMarks
[2]i
3.5b
AnswerMarks
3.5bmust be comments about
motion
AnswerMarks Guidance
16(iii) e
p
e(cid:16)2 (cid:83)k
3 (cid:32)0.98
S
AnswerMarks
k (cid:32)0.009646M1
B1
A1
AnswerMarks
[3]3.1b
3.3
AnswerMarks
1.1Attempt – may use incorrect
period
for correct period
AnswerMarks Guidance
16(iv) Starting with x(cid:32)e(cid:16)kt(Acos3t(cid:14)Bsin3t), with k as in
(iii).
t(cid:32)0,x(cid:32)0 gives A(cid:32)0
dx
(cid:32)e(cid:16)kt(3Bcos3t(cid:16)kBsin3t)
dt
dx
t(cid:32)0, (cid:32)12 gives B(cid:32)4
dt
AnswerMarks
x(cid:32)4e(cid:16)ktsin3twith k as in (iii)B1
M1
A1
B1
AnswerMarks
[4]1.1a
1.1
2.4
AnswerMarks
3.3n
Must see reasoning for where
B comes from.
e
AnswerMarks Guidance
16(v) Without damping ie k (cid:32)0 the amplitude is 4
In fact there is slight damping, so amplitude slightly
AnswerMarks
less than 4.E1
E1
AnswerMarks
[2]m
3.4
3.5a
AnswerMarks Guidance
QuestionAO1 AO2
Question 16:
16 | (i) | (A) | 2(cid:83)
T (cid:32) with T (cid:32)2
(cid:90)
(cid:90)(cid:32)(cid:83)
d2x
(cid:32)(cid:16)(cid:83)2x
dt2 | M1
A1
B1
[3] | 3.1b
1.1
3.3
16 | (i) | (B) | x(cid:32)acos(cid:83)t(cid:14)bsin(cid:83)t | B1
[1] | 1.1 | n
or e.g. x(cid:32) Asin((cid:83)t(cid:14)(cid:72))
16 | (ii) | (A) | Auxiliary equation (cid:79)2(cid:14)2k(cid:79)(cid:14)(k2(cid:14)9)(cid:32)0
Roots (cid:79)(cid:32)(cid:16)k(cid:114)3i
Hence x(cid:32)e(cid:16)kt(Acos3t(cid:14)Bsin3t) | M1
A1
A1
[3] | 1.1a
m
1.1
2.2a | e
16 | (ii) | (B) | Motion is damped [SHM] …
… and period is a bit more than 2s | c
E1
E1
[2] | i
3.5b
3.5b | must be comments about
motion
16 | (iii) | e
p
e(cid:16)2 (cid:83)k
3 (cid:32)0.98
S
k (cid:32)0.009646 | M1
B1
A1
[3] | 3.1b
3.3
1.1 | Attempt – may use incorrect
period
for correct period
16 | (iv) | Starting with x(cid:32)e(cid:16)kt(Acos3t(cid:14)Bsin3t), with k as in
(iii).
t(cid:32)0,x(cid:32)0 gives A(cid:32)0
dx
(cid:32)e(cid:16)kt(3Bcos3t(cid:16)kBsin3t)
dt
dx
t(cid:32)0, (cid:32)12 gives B(cid:32)4
dt
x(cid:32)4e(cid:16)ktsin3twith k as in (iii) | B1
M1
A1
B1
[4] | 1.1a
1.1
2.4
3.3 | n
Must see reasoning for where
B comes from.
e
16 | (v) | Without damping ie k (cid:32)0 the amplitude is 4
In fact there is slight damping, so amplitude slightly
less than 4. | E1
E1
[2] | m
3.4
3.5a
Question | AO1 | AO2 | AO3(PS) | AO3(M) | Totals
A small object is attached to a spring and performs oscillations in a vertical line. The displacement of the object at time $t$ seconds is denoted by $x$ cm.

Preliminary observations suggest that the object performs simple harmonic motion (SHM) with a period of 2 seconds about the point at which $x = 0$.

\begin{enumerate}[label=(\roman*)]
\item \begin{enumerate}[label=(\Alph*)]
\item Write down a differential equation to model this motion. [3]

\item Give the general solution of the differential equation in part (i) (A). [1]
\end{enumerate}
\end{enumerate}

Subsequent observations indicate that the object's motion would be better modelled by the differential equation
$$\frac{d^2x}{dt^2} + 2k \frac{dx}{dt} + (k^2 + 9)x = 0 \qquad (*)$$
where $k$ is a positive constant.

\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item \begin{enumerate}[label=(\Alph*)]
\item Obtain the general solution of (*). [3]

\item State two ways in which the motion given by this model differs from that in part (i). [2]
\end{enumerate}
\end{enumerate}

The amplitude of the object's motion is observed to reduce with a scale factor of 0.98 from one oscillation to the next.

\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item Find the value of $k$. [3]
\end{enumerate}

At the start of the object's motion, $x = 0$ and the velocity is 12 cm s$^{-1}$ in the positive $x$ direction.

\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{3}
\item Find an equation for $x$ as a function of $t$. [4]

\item Without doing any further calculations, explain why, according to this model, the greatest distance of the object from its starting point in the subsequent motion will be slightly less than 4 cm. [2]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Pure Core  Q16 [18]}}