OCR Further Pure Core 2 2022 June — Question 6 10 marks

Exam BoardOCR
ModuleFurther Pure Core 2 (Further Pure Core 2)
Year2022
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSecond order differential equations
TypeModeling context with interpretation
DifficultyChallenging +1.2 This is a structured Further Maths question on second-order differential equations with physical context. Part (a) requires standard recall of SHM solutions, parts (b)-(c) involve applying boundary conditions and algebraic manipulation, part (d) requires integration for a mean value, and parts (e)-(f) involve error analysis and optimization. While it's a multi-part question requiring several techniques, each step is fairly guided and uses standard Further Maths methods without requiring particularly novel insight. The modeling context and error bounds add some complexity above a typical Core Pure question, placing it moderately above average difficulty.
Spec4.08e Mean value of function: using integral4.10f Simple harmonic motion: x'' = -omega^2 x

6 A particle, \(P\), positioned at the origin, \(O\), is projected with a certain velocity along the \(x\)-axis. \(P\) is then acted on by a single force which varies in such a way that \(P\) moves backwards and forwards along the \(x\)-axis. When the time after projection is \(t\) seconds, the displacement of \(P\) from the origin is \(x \mathrm {~m}\) and its velocity is \(v \mathrm {~ms} ^ { - 1 }\). The motion of \(P\) is modelled using the differential equation \(\ddot { x } + \omega ^ { 2 } x = 0\), where \(\omega\) rads \(^ { - 1 }\) is a positive constant.
  1. Write down the general solution of this differential equation. \(D\) is the point where \(x = d\) for some positive constant, \(d\). When \(P\) reaches \(D\) it comes to instantaneous rest.
  2. Using the answer to part (a), determine expressions, in terms of \(\omega\), \(d\) and \(t\) only, for the following quantities
    The quantity \(z\) is defined by \(z = \frac { 1 } { v }\).
  3. Using part (c), determine an expression for \(\mathrm { Z } _ { \mathrm { m } }\), the mean value of z with respect to the displacement, as \(P\) moves directly from \(O\) to \(D\). One measure of the validity of the model is consideration of the value of \(\mathrm { z } _ { \mathrm { m } }\). If \(\mathrm { z } _ { \mathrm { m } }\) exceeds 8 then the model is considered to be valid. The value of \(d\) is measured as 0.25 to 2 significant figures. The value of \(\omega\) is measured as \(0.75 \pm 0.02\).
  4. Determine what can be inferred about the validity of the model from the given information.
  5. Find, according to the model, the least possible value of the velocity with which \(P\) was initially projected. Give your answer to \(\mathbf { 2 }\) significant figures.

Question 6:
AnswerMarks Guidance
6(a) x = Asint + Bcost or Rcos(t + )
or Rsin(t + )B1
[1]1.2 Correct form with 2 arbitrary
constants. Must be “x =”.
Do not ISW; consider final answer
as GS unless explicitly labelled
AnswerMarks
otherwise.Candidates may derive GS from, eg,
auxiliary equation but GS must be in
real form.
AnswerMarks
(b)t = 0, x = 0 => B = 0 (so x = Asint)
Stops when x = d => A = d so x = dsint
AnswerMarks
v = dcostM1
A1
A1
AnswerMarks
[3]3.3
3.4
AnswerMarks
2.2aUsing one boundary condition (may
be seen in (a)).
Using other boundary condition
AnswerMarks
(c)R H S 2 ( 2 d x 2 ) 2 ( d 2 d 2 s i n 2 t )    = − = −
2 d 2 (1 s i n 2 t ) 2 2 d c o s 2 t     = − =
AnswerMarks Guidance
( d c o s 2 t ) v 2 L H S   = = =B1
[1]3.4 AG. Sufficient working must be
shown.
AnswerMarks
(d)1 d0 1 d0 1
z = z d x = d x
m d − 0 d v
1 d 1
d x = 
d d 2 x 2  −
0
d
1 x 1  
s in 1 s in 1 1 s in 1 0 = − =  − − − 
d d d  
0
=
AnswerMarks
2 d M1
A1
AnswerMarks
[2]3.3
3.4Use of mean formula, over x, with
correct limits and z substituted
AnswerMarks
(e)Using d = 0.25 and  = 0.75 leads to z = 8.38...,
m
which suggests that the model is valid, but d could
be as high as 0.255 and  as high as 0.77 which
would lead to z = 7.99998. So it is highly likely
m
that the model is valid (although just possible that it
AnswerMarks Guidance
is not).B1
[1]2.2b Indication that for most, but not all,
of the possible combinations of
values of d and  the value of z
m
AnswerMarks
exceeds 8.7.99998129... < z < 8.782758327...
m
Question 6:
6 | (a) | x = Asint + Bcost or Rcos(t + )
or Rsin(t + ) | B1
[1] | 1.2 | Correct form with 2 arbitrary
constants. Must be “x =”.
Do not ISW; consider final answer
as GS unless explicitly labelled
otherwise. | Candidates may derive GS from, eg,
auxiliary equation but GS must be in
real form.
(b) | t = 0, x = 0 => B = 0 (so x = Asint)
Stops when x = d => A = d so x = dsint
v = dcost | M1
A1
A1
[3] | 3.3
3.4
2.2a | Using one boundary condition (may
be seen in (a)).
Using other boundary condition
(c) | R H S 2 ( 2 d x 2 ) 2 ( d 2 d 2 s i n 2 t )    = − = −
2 d 2 (1 s i n 2 t ) 2 2 d c o s 2 t     = − =
( d c o s 2 t ) v 2 L H S   = = = | B1
[1] | 3.4 | AG. Sufficient working must be
shown.
(d) | 1 d0 1 d0 1
z = z d x = d x
m d − 0 d v
1 d 1
d x = 
d d 2 x 2  −
0
d
1 x 1  
s in 1 s in 1 1 s in 1 0 = − =  − − − 
d d d  
0

=
2 d  | M1
A1
[2] | 3.3
3.4 | Use of mean formula, over x, with
correct limits and z substituted
(e) | Using d = 0.25 and  = 0.75 leads to z = 8.38...,
m
which suggests that the model is valid, but d could
be as high as 0.255 and  as high as 0.77 which
would lead to z = 7.99998. So it is highly likely
m
that the model is valid (although just possible that it
is not). | B1
[1] | 2.2b | Indication that for most, but not all,
of the possible combinations of
values of d and  the value of z
m
exceeds 8. | 7.99998129... < z < 8.782758327...
m
6 A particle, $P$, positioned at the origin, $O$, is projected with a certain velocity along the $x$-axis. $P$ is then acted on by a single force which varies in such a way that $P$ moves backwards and forwards along the $x$-axis.

When the time after projection is $t$ seconds, the displacement of $P$ from the origin is $x \mathrm {~m}$ and its velocity is $v \mathrm {~ms} ^ { - 1 }$.

The motion of $P$ is modelled using the differential equation $\ddot { x } + \omega ^ { 2 } x = 0$, where $\omega$ rads $^ { - 1 }$ is a positive constant.
\begin{enumerate}[label=(\alph*)]
\item Write down the general solution of this differential equation.\\
$D$ is the point where $x = d$ for some positive constant, $d$. When $P$ reaches $D$ it comes to instantaneous rest.
\item Using the answer to part (a), determine expressions, in terms of $\omega$, $d$ and $t$ only, for the following quantities

\begin{itemize}
  \item $X$
  \item $v$
\item Hence show that, according to the model, $v ^ { 2 } = \omega ^ { 2 } \left( d ^ { 2 } - x ^ { 2 } \right)$.
\end{itemize}

The quantity $z$ is defined by $z = \frac { 1 } { v }$.
\item Using part (c), determine an expression for $\mathrm { Z } _ { \mathrm { m } }$, the mean value of z with respect to the displacement, as $P$ moves directly from $O$ to $D$.

One measure of the validity of the model is consideration of the value of $\mathrm { z } _ { \mathrm { m } }$. If $\mathrm { z } _ { \mathrm { m } }$ exceeds 8 then the model is considered to be valid.

The value of $d$ is measured as 0.25 to 2 significant figures. The value of $\omega$ is measured as $0.75 \pm 0.02$.
\item Determine what can be inferred about the validity of the model from the given information.
\item Find, according to the model, the least possible value of the velocity with which $P$ was initially projected. Give your answer to $\mathbf { 2 }$ significant figures.
\end{enumerate}

\hfill \mbox{\textit{OCR Further Pure Core 2 2022 Q6 [10]}}