5.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{44066c44-e366-4f87-b1b2-c5a894e407fa-16_193_1367_274_287}
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\caption{Figure 2}
\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, \(A P = ( 3 + x ) \mathrm { m }\).
- 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.
- Find the time at which \(P\) first comes to instantaneous rest.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{44066c44-e366-4f87-b1b2-c5a894e407fa-20_705_1104_116_420}
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\caption{Figure 3}
\end{figure}