Edexcel M4 2017 June — Question 6 13 marks

Exam BoardEdexcel
ModuleM4 (Mechanics 4)
Year2017
SessionJune
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSecond order differential equations
TypeParticular solution with initial conditions
DifficultyStandard +0.8 This is a damped harmonic motion problem requiring derivation of the differential equation from forces, solving a second-order DE with complex roots (underdamped case), applying initial conditions, and finding when velocity equals zero. While methodical, it involves multiple substantial steps across mechanics and differential equations, placing it moderately above average difficulty for A-level.
Spec4.10g Damped oscillations: model and interpret6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle

6. A particle \(P\) of mass 0.2 kg is suspended from a fixed point by a light elastic spring. The spring has natural length 0.8 m and modulus of elasticity 7 N . At time \(t = 0\) the particle is released from rest from a point 0.2 metres vertically below its equilibrium position. The motion of \(P\) is resisted by a force of magnitude \(2 v\) newtons, where \(v \mathrm {~ms} ^ { - 1 }\) is the speed of \(P\). At time \(t\) seconds, \(P\) is \(x\) metres below its equilibrium position.
  1. Show that \(\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 10 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 43.75 x = 0\)
  2. Find \(x\) in terms of \(t\).
  3. Find the value of \(t\) when \(P\) first comes to instantaneous rest.

6. A particle $P$ of mass 0.2 kg is suspended from a fixed point by a light elastic spring. The spring has natural length 0.8 m and modulus of elasticity 7 N . At time $t = 0$ the particle is released from rest from a point 0.2 metres vertically below its equilibrium position. The motion of $P$ is resisted by a force of magnitude $2 v$ newtons, where $v \mathrm {~ms} ^ { - 1 }$ is the speed of $P$. At time $t$ seconds, $P$ is $x$ metres below its equilibrium position.
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 10 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 43.75 x = 0$
\item Find $x$ in terms of $t$.
\item Find the value of $t$ when $P$ first comes to instantaneous rest.
\end{enumerate}

\hfill \mbox{\textit{Edexcel M4 2017 Q6 [13]}}