| Exam Board | OCR MEI |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2010 |
| Session | June |
| Marks | 18 |
| 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 a straightforward differentiation exercise to verify SHM (2 differentiations). Part (ii) involves routine substitution of initial conditions into three equations. Parts (iii)-(v) require standard SHM formulas and understanding of amplitude/period, but are mostly procedural applications. This is a typical M3/mechanics module question testing standard SHM techniques with no novel insight required, making it slightly easier than average overall. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(v = \frac{dx}{dt} = A\omega\cos\omega t - B\omega\sin\omega t\) | B1 | |
| \(a = \frac{d^2x}{dt^2} = -A\omega^2\sin\omega t - B\omega^2\cos\omega t\) | M1 | Finding the second derivative |
| \(= -\omega^2(A\sin\omega t + B\cos\omega t) = -\omega^2 x\) | E1 | Correctly shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(B = -16\) | B1 | |
| \(\omega = 0.25\) | B1 | |
| \(A = 30\) | B2 | When \(A\) is wrong, give B1 for a correct equation involving \(A\) [e.g. \(A\omega = 7.5\) or \(7.5^2 = \omega^2(A^2 + B^2 - 16^2)\)] or for \(A = -30\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Maximum displacement is \((\pm)\sqrt{A^2 + B^2}\) | M1 | Or \(7.5^2 = \omega^2(\text{amp}^2 - 16^2)\); or finding \(t\) when \(v=0\) and substituting to find \(x\) |
| Maximum displacement is \(34\) m | A1 | |
| Maximum speed is \((\pm)34\omega\) | M1 | For either (any valid method) |
| Maximum acceleration is \((\pm)34\omega^2\) | ||
| Maximum speed is \(8.5\ \text{ms}^{-1}\) | F1 | Only ft from \(\omega \times \text{amp}\) |
| Maximum acceleration is \(2.125\ \text{ms}^{-2}\) | F1 | Only ft from \(\omega^2 \times \text{amp}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(v = 7.5\cos 0.25t + 4\sin 0.25t\) | ||
| When \(t=15\): \(v = 7.5\cos 3.75 + 4\sin 3.75 = -8.44\) | M1 | |
| Speed is \(8.44\ \text{ms}^{-1}\) (3 sf); downwards | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Period \(\frac{2\pi}{\omega} \approx 25\) s, so \(t=0\) to \(t=15\) is less than one period | ||
| When \(t=15\): \(x = 30\sin 3.75 - 16\cos 3.75 = -4.02\) | M1 | |
| Distance travelled is \(16 + 34 + 34 + 4.02\) | M1 | Take account of change of direction |
| M1 | Fully correct strategy for distance | |
| Distance travelled is \(88.0\) m (3 sf) | A1 cao |
# Question 4:
## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $v = \frac{dx}{dt} = A\omega\cos\omega t - B\omega\sin\omega t$ | B1 | |
| $a = \frac{d^2x}{dt^2} = -A\omega^2\sin\omega t - B\omega^2\cos\omega t$ | M1 | Finding the second derivative |
| $= -\omega^2(A\sin\omega t + B\cos\omega t) = -\omega^2 x$ | E1 | Correctly shown |
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $B = -16$ | B1 | |
| $\omega = 0.25$ | B1 | |
| $A = 30$ | B2 | When $A$ is wrong, give B1 for a correct equation involving $A$ [e.g. $A\omega = 7.5$ or $7.5^2 = \omega^2(A^2 + B^2 - 16^2)$] or for $A = -30$ |
## Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Maximum displacement is $(\pm)\sqrt{A^2 + B^2}$ | M1 | Or $7.5^2 = \omega^2(\text{amp}^2 - 16^2)$; or finding $t$ when $v=0$ and substituting to find $x$ |
| Maximum displacement is $34$ m | A1 | |
| Maximum speed is $(\pm)34\omega$ | M1 | For either (any valid method) |
| Maximum acceleration is $(\pm)34\omega^2$ | | |
| Maximum speed is $8.5\ \text{ms}^{-1}$ | F1 | Only ft from $\omega \times \text{amp}$ |
| Maximum acceleration is $2.125\ \text{ms}^{-2}$ | F1 | Only ft from $\omega^2 \times \text{amp}$ |
## Part (iv)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $v = 7.5\cos 0.25t + 4\sin 0.25t$ | | |
| When $t=15$: $v = 7.5\cos 3.75 + 4\sin 3.75 = -8.44$ | M1 | |
| Speed is $8.44\ \text{ms}^{-1}$ (3 sf); downwards | A1 | |
## Part (v)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Period $\frac{2\pi}{\omega} \approx 25$ s, so $t=0$ to $t=15$ is less than one period | | |
| When $t=15$: $x = 30\sin 3.75 - 16\cos 3.75 = -4.02$ | M1 | |
| Distance travelled is $16 + 34 + 34 + 4.02$ | M1 | Take account of change of direction |
| | M1 | Fully correct strategy for distance |
| Distance travelled is $88.0$ m (3 sf) | A1 cao | |4 A particle P is performing simple harmonic motion in a vertical line. At time $t \mathrm {~s}$, its displacement $x \mathrm {~m}$ above a fixed point O is given by
$$x = A \sin \omega t + B \cos \omega t$$
where $A , B$ and $\omega$ are constants.\\
(i) Show that the acceleration of P , in $\mathrm { ms } ^ { - 2 }$, is $- \omega ^ { 2 } x$.
When $t = 0 , \mathrm { P }$ is 16 m below O , moving with velocity $7.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ upwards, and has acceleration $1 \mathrm {~m} \mathrm {~s} ^ { - 2 }$ upwards.\\
(ii) Find the values of $A , B$ and $\omega$.\\
(iii) Find the maximum displacement, the maximum speed, and the maximum acceleration of P .\\
(iv) Find the speed and the direction of motion of P when $t = 15$.\\
(v) Find the distance travelled by P between $t = 0$ and $t = 15$.
\hfill \mbox{\textit{OCR MEI M3 2010 Q4 [18]}}