| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2015 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Time to travel between positions |
| Difficulty | Challenging +1.2 This is a standard Further Maths SHM problem requiring systematic application of the standard formulas (a = -ω²x, v² = ω²(a² - x²)) to find period and amplitude, followed by integration to find time between positions. While it involves multiple steps and careful setup, the techniques are routine for FP2 students with no novel insight required beyond recognizing which formulas to apply. |
| Spec | 1.05g Exact trigonometric values: for standard angles4.10f Simple harmonic motion: x'' = -omega^2 x |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(0.625 = 10\omega^2\), \(\omega^2 = 0.0625\) or \(1/16\) | B1 | Find \(\omega^2\) from SHM eqn. \(d^2x/dt^2 = -\omega^2 x\) at \(C\) |
| \(T = 2\pi/\frac{1}{4} = 8\pi\) (not 25.1) | B1\(\checkmark\) | Find period \(T\) [s] from \(T = 2\pi/\omega\) (ft on \(\omega^2\)) |
| Total: 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(6^2 = \omega^2(a^2 - 10^2)\) | Find amplitude \(a\) [m] from \(v_C^2 = \omega^2(a^2 - 10^2)\) | |
| \(a^2 = 6^2 \times 16 + 10^2\), \(a = \sqrt{676} = 26\) | M1 A1 | |
| \(\omega^{-1}\sin^{-1}(10/a) + \omega^{-1}\sin^{-1}\frac{1}{2}\) *or* | Find time from \(C\) to \(M\) | |
| \(\omega^{-1}\cos^{-1}(-10/a) - \omega^{-1}\cos^{-1}\frac{1}{2}\) *or* | ||
| \(\frac{1}{2}T - \omega^{-1}\cos^{-1}(10/a) - \omega^{-1}\cos^{-1}\frac{1}{2}\) | M1 | |
| \(= \omega^{-1}\{0.3948 + \pi/6\ [= 0.5236]\}\) *or* | ||
| \(\omega^{-1}\{1.9656 - \pi/3\ [= 1.0472]\}\) *or* | ||
| \(\omega^{-1}\{\pi - 1.760 - \pi/3\ [= 1.0472]\}\) | A1 | |
| \(= 1.579 + 2.094\) *or* \(7.862 - 4.189\) | ||
| *or* \(12.567 - 4.704 - 4.189\) | ||
| *or* \(4 \times 0.9184 = 3.67\) [s] | A1; A1 | AEF throughout |
| Total: 4 marks |
## Question 2(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $0.625 = 10\omega^2$, $\omega^2 = 0.0625$ or $1/16$ | B1 | Find $\omega^2$ from SHM eqn. $d^2x/dt^2 = -\omega^2 x$ at $C$ |
| $T = 2\pi/\frac{1}{4} = 8\pi$ (not 25.1) | B1$\checkmark$ | Find period $T$ [s] from $T = 2\pi/\omega$ (ft on $\omega^2$) |
| **Total: 2 marks** | | |
## Question 2(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $6^2 = \omega^2(a^2 - 10^2)$ | | Find amplitude $a$ [m] from $v_C^2 = \omega^2(a^2 - 10^2)$ |
| $a^2 = 6^2 \times 16 + 10^2$, $a = \sqrt{676} = 26$ | M1 A1 | |
| $\omega^{-1}\sin^{-1}(10/a) + \omega^{-1}\sin^{-1}\frac{1}{2}$ *or* | | Find time from $C$ to $M$ |
| $\omega^{-1}\cos^{-1}(-10/a) - \omega^{-1}\cos^{-1}\frac{1}{2}$ *or* | | |
| $\frac{1}{2}T - \omega^{-1}\cos^{-1}(10/a) - \omega^{-1}\cos^{-1}\frac{1}{2}$ | M1 | |
| $= \omega^{-1}\{0.3948 + \pi/6\ [= 0.5236]\}$ *or* | | |
| $\omega^{-1}\{1.9656 - \pi/3\ [= 1.0472]\}$ *or* | | |
| $\omega^{-1}\{\pi - 1.760 - \pi/3\ [= 1.0472]\}$ | A1 | |
| $= 1.579 + 2.094$ *or* $7.862 - 4.189$ | | |
| *or* $12.567 - 4.704 - 4.189$ | | |
| *or* $4 \times 0.9184 = 3.67$ [s] | A1; A1 | AEF throughout |
| **Total: 4 marks** | | |2 A particle $P$ moves on a straight line $A O B$ in simple harmonic motion. The centre of the motion is $O$, and $P$ is instantaneously at rest at $A$ and $B$. The point $C$ is on the line $A O B$, between $A$ and $O$, and $C O = 10 \mathrm {~m}$. When $P$ is at $C$, the magnitude of its acceleration is $0.625 \mathrm {~m} \mathrm {~s} ^ { - 2 }$ and it is moving towards $O$ with speed $6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Find\\
(i) the period of the motion, in terms of $\pi$,\\
(ii) the amplitude of the motion.
The point $M$ is the mid-point of $O B$. Find the time that $P$ takes to travel directly from $C$ to $M$.
\hfill \mbox{\textit{CAIE FP2 2015 Q2 [8]}}