| Exam Board | OCR MEI |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2008 |
| Session | January |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Prove motion is SHM from equation |
| Difficulty | Standard +0.3 Part (i) is routine differentiation to verify SHM. Parts (ii-iii) involve standard initial condition substitutions and formula application. Part (iv) requires understanding of periodic motion to calculate distance over multiple cycles, which adds modest problem-solving beyond pure recall. Overall slightly above average difficulty due to the multi-part nature and final distance calculation, but all techniques are standard M3 material. |
| Spec | 1.07d Second derivatives: d^2y/dx^2 notation4.10b Model with differential equations: kinematics and other contexts4.10f Simple harmonic motion: x'' = -omega^2 x |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dx}{dt} = A\omega\cos\omega t - B\omega\sin\omega t\) | B1 | |
| \(\frac{d^2x}{dt^2} = -A\omega^2\sin\omega t - B\omega^2\cos\omega t\) | B1 ft | Must follow from their \(\dot{x}\) |
| \(= -\omega^2(A\sin\omega t + B\cos\omega t) = -\omega^2 x\) | E1 | [3] Fully correct completion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(B = 2\) | B1 | |
| \(A\omega = -1.44\) | M1 | Using \(\frac{dx}{dt} = -1.44\) when \(t = 0\) |
| A1 cao | ||
| \(-B\omega^2 = -0.18\) or \(-0.18 = -\omega^2(2)\) | M1 | \(\frac{d^2x}{dt^2} = -0.18\) when \(t=0\) (or \(x=2\)) |
| A1 cao | ||
| \(\omega = 0.3\), \(\quad A = -4.8\) | A1 cao | [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Period is \(\frac{2\pi}{\omega} = \frac{2\pi}{0.3} = 20.94 = 20.9 \text{ s}\) (3 sf) | E1 | or \(1.44^2 = 0.3^2(a^2 - 2^2)\) |
| Amplitude is \(\sqrt{A^2 + B^2} = \sqrt{4.8^2 + 2^2}\) | M1 | |
| \(= 5.2 \text{ m}\) | A1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x = -4.8\sin 0.3t + 2\cos 0.3t\) | ||
| \(v = -1.44\cos 0.3t - 0.6\sin 0.3t\) | ||
| When \(t=12\): \(x = 0.3306\) \((v = 1.56)\) | M1 | Finding \(x\) when \(t=12\) and \(t=24\) |
| When \(t=24\): \(x = -2.5929\) \((v = -1.35)\) | A1 | Both displacements correct |
| Distance travelled is \((5.2 - 0.3306) + 5.2 + 2.5929 = 12.7 \text{ m}\) | M1 M1 A1 | [5] Considering change of direction; Correct method for distance |
# Question 3:
## Part 3(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dx}{dt} = A\omega\cos\omega t - B\omega\sin\omega t$ | B1 | |
| $\frac{d^2x}{dt^2} = -A\omega^2\sin\omega t - B\omega^2\cos\omega t$ | B1 ft | Must follow from their $\dot{x}$ |
| $= -\omega^2(A\sin\omega t + B\cos\omega t) = -\omega^2 x$ | E1 | **[3]** Fully correct completion |
*SR: For $\dot{x} = -A\omega\cos\omega t + B\omega\sin\omega t$, $\ddot{x} = -A\omega^2\sin\omega t - B\omega^2\cos\omega t$, award B0B1E0*
## Part 3(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $B = 2$ | B1 | |
| $A\omega = -1.44$ | M1 | Using $\frac{dx}{dt} = -1.44$ when $t = 0$ |
| | A1 cao | |
| $-B\omega^2 = -0.18$ or $-0.18 = -\omega^2(2)$ | M1 | $\frac{d^2x}{dt^2} = -0.18$ when $t=0$ (or $x=2$) |
| | A1 cao | |
| $\omega = 0.3$, $\quad A = -4.8$ | A1 cao | **[6]** |
## Part 3(iii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Period is $\frac{2\pi}{\omega} = \frac{2\pi}{0.3} = 20.94 = 20.9 \text{ s}$ (3 sf) | E1 | or $1.44^2 = 0.3^2(a^2 - 2^2)$ |
| Amplitude is $\sqrt{A^2 + B^2} = \sqrt{4.8^2 + 2^2}$ | M1 | |
| $= 5.2 \text{ m}$ | A1 | **[3]** |
## Part 3(iv)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x = -4.8\sin 0.3t + 2\cos 0.3t$ | | |
| $v = -1.44\cos 0.3t - 0.6\sin 0.3t$ | | |
| When $t=12$: $x = 0.3306$ $(v = 1.56)$ | M1 | Finding $x$ when $t=12$ and $t=24$ |
| When $t=24$: $x = -2.5929$ $(v = -1.35)$ | A1 | Both displacements correct |
| Distance travelled is $(5.2 - 0.3306) + 5.2 + 2.5929 = 12.7 \text{ m}$ | M1 M1 A1 | **[5]** Considering change of direction; Correct method for distance |
*ft from their $A, B, \omega$ and amplitude. Third M1 requires method comparable to correct one. A1A1 both require $\omega \approx 0.3$, $A \neq 0$, $B \neq 0$*
*Note: ft from $A = +4.8$: $x_{12} = -3.92$ $(v<0)$, $x_{24} = 5.03$ $(v>0)$. Distance is $(5.2-3.92)+5.2+5.03 = 11.5$*
---3 A particle is oscillating in a vertical line. At time $t$ seconds, its displacement above the centre of the oscillations is $x$ metres, where $x = A \sin \omega t + B \cos \omega t$ (and $A , B$ and $\omega$ are constants).\\
(i) Show that $\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } = - \omega ^ { 2 } x$.
When $t = 0$, the particle is 2 m above the centre of the oscillations, the velocity is $1.44 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ downwards, and the acceleration is $0.18 \mathrm {~m} \mathrm {~s} ^ { - 2 }$ downwards.\\
(ii) Find $A , B$ and $\omega$.\\
(iii) Show that the period of oscillation is 20.9 s (correct to 3 significant figures), and find the amplitude.\\
(iv) Find the total distance travelled by the particle between $t = 12$ and $t = 24$.
\hfill \mbox{\textit{OCR MEI M3 2008 Q3 [17]}}