OCR MEI C4 2013 June — Question 6 18 marks

Exam BoardOCR MEI
ModuleC4 (Core Mathematics 4)
Year2013
SessionJune
Marks18
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHarmonic Form
TypeFind value where max/min occurs
DifficultyStandard +0.3 This is a standard C4 differential equation question with routine techniques: separating variables (part i), verifying initial conditions (part ii), using R cos(θ-α) form and differentiation (part iii), and finding maxima (part iv). All parts follow textbook methods with no novel insight required, though the multi-part structure and algebraic manipulation make it slightly above average difficulty for typical A-level questions.
Spec1.05l Double angle formulae: and compound angle formulae1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.08k Separable differential equations: dy/dx = f(x)g(y)4.10a General/particular solutions: of differential equations4.10b Model with differential equations: kinematics and other contexts4.10f Simple harmonic motion: x'' = -omega^2 x

The motion of a particle is modelled by the differential equation $$v \frac{dv}{dt} + 4x = 0,$$ where \(x\) is its displacement from a fixed point, and \(v\) is its velocity. Initially \(x = 1\) and \(v = 4\).
  1. Solve the differential equation to show that \(v^2 = 20 - 4x^2\). [4]
Now consider motion for which \(x = \cos 2t + 2 \sin 2t\), where \(x\) is the displacement from a fixed point at time \(t\).
  1. Verify that, when \(t = 0\), \(x = 1\). Use the fact that \(v = \frac{dx}{dt}\) to verify that when \(t = 0\), \(v = 4\). [4]
  2. Express \(x\) in the form \(R \cos(2t - \alpha)\), where \(R\) and \(\alpha\) are constants to be determined, and obtain the corresponding expression for \(v\). Hence or otherwise verify that, for this motion too, \(v^2 = 20 - 4x^2\). [7]
  3. Use your answers to part (iii) to find the maximum value of \(x\), and the earliest time at which \(x\) reaches this maximum value. [3]

The motion of a particle is modelled by the differential equation

$$v \frac{dv}{dt} + 4x = 0,$$

where $x$ is its displacement from a fixed point, and $v$ is its velocity.

Initially $x = 1$ and $v = 4$.

\begin{enumerate}[label=(\roman*)]
\item Solve the differential equation to show that $v^2 = 20 - 4x^2$. [4]
\end{enumerate}

Now consider motion for which $x = \cos 2t + 2 \sin 2t$, where $x$ is the displacement from a fixed point at time $t$.

\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Verify that, when $t = 0$, $x = 1$. Use the fact that $v = \frac{dx}{dt}$ to verify that when $t = 0$, $v = 4$. [4]

\item Express $x$ in the form $R \cos(2t - \alpha)$, where $R$ and $\alpha$ are constants to be determined, and obtain the corresponding expression for $v$. Hence or otherwise verify that, for this motion too, $v^2 = 20 - 4x^2$. [7]

\item Use your answers to part (iii) to find the maximum value of $x$, and the earliest time at which $x$ reaches this maximum value. [3]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI C4 2013 Q6 [18]}}