| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2015 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Displacement and velocity at given time |
| Difficulty | Standard +0.3 This is a standard M3 SHM question requiring application of standard formulas (x = a cos(ωt + ε), v = -aω sin(ωt + ε)) with straightforward initial conditions. Part (i) involves finding amplitude and ω from given information, part (ii) is direct substitution, and part (iii) requires understanding that equal speeds occur at symmetric positions, which is a common SHM concept. The multi-part structure and 11 marks indicate moderate length, but no novel insight or complex problem-solving is required beyond textbook techniques. |
| Spec | 3.02g Two-dimensional variable acceleration4.10f Simple harmonic motion: x'' = -omega^2 x |
| Answer | Marks |
|---|---|
| B1 | |
| B1 | |
| B1f[3] | accept sight of \(\frac{\pi}{0.25s}\) or \(\frac{2\pi}{0.5s}\); fl from wrong \(a\) and/or \(\omega\) |
| Answer | Marks |
|---|---|
| M1 | use of \(a\cos\omega t\); complete method |
| A1 | |
| M1 | use of \((-) a\omega\sin\omega t\) or \(v^2 = \omega^2(a^2 - x^2)\); allow M1fl from wrong formula for \(x\) \(-0.80397......\) |
| A1 [4] | if \(v^2\) formula used, direction of \(v\) needs to be made clear. |
| Answer | Marks |
|---|---|
| B2 | values of \(t\) are = 0.0854, 0.871 |
| B1 | values of \(x\) are 0.565, -0.565 |
| B1 | dep first 3 marks |
| ignore ref to point when \(t = 0.7\) can show on diagram | |
| [4] | ignore wrong values |
### Part (i)
**Answer:** $a = 0.6$ (m)
$\omega = 4$
max vel = $a\omega = 2.4$ (m s$^{-1}$)
| B1 | |
| B1 | |
| B1f[3] | accept sight of $\frac{\pi}{0.25s}$ or $\frac{2\pi}{0.5s}$; fl from wrong $a$ and/or $\omega$ |
### Part (ii)
**Answer:** must use their $a$ and $\omega$ from (i) unless defined differently in (ii)
$x = 0.6\cos 4x0.7$
$x = -0.565$
$v = -0.6x4\sin4x0.7$
$v = -0.804$
| M1 | use of $a\cos\omega t$; complete method |
| A1 | |
| M1 | use of $(-) a\omega\sin\omega t$ or $v^2 = \omega^2(a^2 - x^2)$; allow M1fl from wrong formula for $x$ $-0.80397......$ |
| A1 [4] | if $v^2$ formula used, direction of $v$ needs to be made clear. |
**Guidance:** or $a\sin(\omega t + c)$, with $c = \pm \pi/2$ $-0.565333...$ or $(-)\omega\cos(\omega t + c)$, with $c = \pm \pi/2$; allow M1fl from wrong formula for $x$ $-0.80397......$
### Part (iii)
**Answer:** do not accept answers from wrong working
$t$ and $x$ for one point
$t$ and $x$ for second point
correctly giving precisely 2 other occasions, with $x$ and $t$ values matching
sc, if < 3 scored, both $t$ values B2
or one $t$ value B1
or $x = 0.565$ B1
of B0 scored allow B1 for number of other occasions shown to be 2
| B2 | values of $t$ are = 0.0854, 0.871 |
| B1 | values of $x$ are 0.565, -0.565 |
| B1 | dep first 3 marks |
| | ignore ref to point when $t = 0.7$ can show on diagram |
| | |
| [4] | ignore wrong values |
**Guidance:** $\pi/4 -0.7$, $\pi/2 - 0.7$; ignore ref to point when $t = 0.7$; can show on diagram
P has this speed 4 times in 1 period (1.570 s) so 2 other times in $0 < t < 1$A particle $P$ starts from rest from a point $A$ and moves in a straight line with simple harmonic motion about a point $O$. At time $t$ seconds after the motion starts the displacement of $P$ from $O$ is $x$ m towards $A$. The particle $P$ is next at rest when $t = 0.25\pi$ having travelled a distance of $1.2$ m.
\begin{enumerate}[label=(\roman*)]
\item Find the maximum velocity of $P$. [3]
\item Find the value of $x$ and the velocity of $P$ when $t = 0.7$. [4]
\item Find the other values of $t$, for $0 < t < 1$, at which $P$'s speed is the same as when $t = 0.7$. Find also the corresponding values of $x$. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR M3 2015 Q6 [11]}}