| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2015 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Collision/impulse during SHM |
| Difficulty | Challenging +1.8 This M4 question combines SHM with a moving boundary condition, requiring careful force analysis with two spring sections and solving a driven harmonic oscillator. The setup is non-standard (particle at midpoint, moving end), demanding geometric reasoning to establish the differential equation. Part (b) requires finding constants from initial conditions and solving a transcendental equation. While the solution is provided, deriving the equation of motion and handling the particular integral represents significant problem-solving beyond routine SHM exercises. |
| Spec | 4.10e Second order non-homogeneous: complementary + particular integral6.02g Hooke's law: T = k*x or T = lambda*x/l6.02h Elastic PE: 1/2 k x^2 |
A particle P of mass 1.5 kg is attached to the midpoint of a light elastic spring AB, of natural length 2 m and modulus of elasticity 12 N. The end A of the spring is attached to a fixed point on a smooth horizontal floor. The end B is held at a point on the floor where $AB = 6$ m.
At time $t = 0$, P is at rest on the floor at the point O, where $AO = 3$ m. The end B is now moved along the floor in such a way that AOB remains a straight line and at time $t$ seconds, $t \geq 0$,
$$AB = \left(6 + \frac{1}{4}\sin 2t\right) \text{ m}$$
At time $t$ seconds, $AP = (3 + x)$ m.
(a) Show that, for $t \geq 0$,
$$\frac{d^2x}{dt^2} + 16x = 2\sin 2t$$
(5 marks)
The general solution of this differential equation is
$$x = C\cos 4t + D\sin 4t + \frac{1}{6}\sin 2t$$
where C and D are constants.
(b) Find the time at which P first comes to instantaneous rest.
(5 marks)
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{44066c44-e366-4f87-b1b2-c5a894e407fa-16_193_1367_274_287}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A particle $P$ of mass 1.5 kg is attached to the midpoint of a light elastic spring $A B$, of natural length 2 m and modulus of elasticity 12 N . The end $A$ of the spring is attached to a fixed point on a smooth horizontal floor. The end $B$ is held at a point on the floor where $A B = 6 \mathrm {~m}$.
At time $t = 0 , P$ is at rest on the floor at the point $O$, where $A O = 3 \mathrm {~m}$, as shown in Figure 2. The end $B$ is now moved along the floor in such a way that $A O B$ remains a straight line and at time $t$ seconds, $t \geqslant 0$,
$$A B = \left( 6 + \frac { 1 } { 4 } \sin 2 t \right) \mathrm { m }$$
At time $t$ seconds, $A P = ( 3 + x ) \mathrm { m }$.
\begin{enumerate}[label=(\alph*)]
\item Show that, for $t \geqslant 0$,
$$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 16 x = 2 \sin 2 t$$
The general solution of this differential equation is
$$x = C \cos 4 t + D \sin 4 t + \frac { 1 } { 6 } \sin 2 t$$
where $C$ and $D$ are constants.
\item Find the time at which $P$ first comes to instantaneous rest.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{44066c44-e366-4f87-b1b2-c5a894e407fa-20_705_1104_116_420}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 2015 Q5 [10]}}