M3 June 2012 Q2
2. A particle \(P\) moves in a straight line with simple harmonic motion about a fixed centre \(O\). The period of the motion is \(\dfrac{\pi}{2}\) seconds. At time \(t\) seconds the speed of \(P\) is \(v\) m s\(^{-1}\). When \(t = 0\), \(P\) is at \(O\) and \(v = 6\). Find
(a) the greatest distance of \(P\) from \(O\) during the motion, (3)
(b) the greatest magnitude of the acceleration of \(P\) during the motion, (2)
(c) the smallest positive value of \(t\) for which \(P\) is 1 m from \(O\). (3)
| Scheme | Marks |
|---|---|
| \(T = \dfrac{2\pi}{\omega} \Rightarrow \omega = 4\) | B1 |
| Use of \(v^2 = \omega^2\left(a^2 - x^2\right)\), or \(v = a\omega\) | M1 |
| \(a = 1.5\) (m) | A1 |
| (3) |
Notes
(Corrected from the printed mark scheme: the formula is printed as \(v^2 = \omega^2\left(v^2 - x^2\right)\).)
| Scheme | Marks |
|---|---|
| Use of max. accn. \(= \omega^2 a\) | M1 |
| 24 m s\(^{-2}\) | A1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(x = a\sin\omega t\) with their values for \(a\) & \(\omega\) | B1 |
| \(1 = 1.5\sin 4t\) (with their 1.5 & 4) and attempt to solve for \(t\) | M1 |
| \(t = 0.18\) (or awrt) | A1 |
| (3) | |
| (8 marks) |