M3 January 2013 Q5
5. A particle \(P\) is moving in a straight line with simple harmonic motion on a smooth horizontal floor. The particle comes to instantaneous rest at points \(A\) and \(B\) where \(AB\) is 0.5 m. The mid-point of \(AB\) is \(O\). The mid-point of \(OA\) is \(C\). The mid-point of \(OB\) is \(D\). The particle takes 0.2 s to travel directly from \(C\) to \(D\). At time \(t = 0\), \(P\) is moving through \(O\) towards \(A\).
(a) Show that the period of the motion is \(\dfrac{6}{5}\) s. (5)
(b) Find the distance of \(P\) from \(B\) when \(t = 2\) s. (3)
(c) Find the maximum magnitude of the acceleration of \(P\). (2)
(d) Find the maximum speed of \(P\). (2)
| Scheme | Marks |
|---|---|
| \(x = a\sin\omega t\) | |
| \(0.125 = 0.25\sin 0.1\omega\) | M1A1 |
| \(\sin 0.1\omega = \dfrac{1}{2}\) | |
| \(0.1\omega = \dfrac{\pi}{6}\) | |
| \(\omega = \dfrac{\pi}{0.6} = \dfrac{10\pi}{6}\) | M1depA1 |
| Period \(= \dfrac{2\pi}{\omega} = \dfrac{6}{5}\ \ \ (= 1.2)\) | A1 (B1 on e-pen) |
| Scheme | Marks |
|---|---|
| \(x = 0.25\sin\dfrac{5}{3}\pi t\) | |
| \(t = 2\ \ \ \ x = 0.25\sin\left(2 \times \dfrac{5}{3}\pi\right)\) | M1 |
| \(x = -0.2165\ldots\) | A1 |
| Dist from \(B = 0.25 + x = 0.033\) m | A1 ft |
| Scheme | Marks |
|---|---|
| Max accel \(= a\omega^2 = 0.25 \times \left(\dfrac{5\pi}{3}\right)^2 = 6.853\ldots = 6.85\) | M1A1 |
| Scheme | Marks |
|---|---|
| Max speed \(a\omega = 0.25 \times \left(\dfrac{5\pi}{3}\right) = 1.308\ldots = 1.31\) | M1A1 |