| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2011 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Prove SHM and find period: vertical spring/string (single attachment) |
| Difficulty | Standard +0.3 This is a standard M3 SHM question requiring Hooke's law equilibrium (routine), showing SHM via F=ma with elastic forces (standard technique), and solving the SHM equation with given initial conditions. All steps follow textbook methods with no novel insight required, but the multi-part structure and calculus/trigonometry involved make it slightly above average difficulty. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| \(2.45e/0.5 = 0.05g\) | M1 | For using \(T = \lambda e/L\) and resolving forces vertically; accept use of 0.1 to show both sides equal to 0.49 |
| \((c = 0.1)\) | A1 | |
| Distance from \(O\) is \(0.5 + 0.1 = 0.6m\) | A1 | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| \(mg - T = m\ddot{x}\) | M1 | For using Newton's second law with 3 terms |
| \(0.05g - 2.45(0.1 + x)/0.5 = 0.05\ddot{x}\) | A1 | |
| \(\ddot{x} = -98x\) | A1 | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| \(a = 0.075\) | B1 | |
| \(n = 7\sqrt{2}\) oe | B1 | accept 9.90 |
| \(x = 0.075\cos(7\sqrt{2} t)\) | M1 | For using \(x = a\cos nt\) oe |
| \(x(0.2) = -0.0298\) | A1 | |
| \(v = -0.075(7\sqrt{2})\sin(7\sqrt{2} t)\) | M1 | For differentiating \(x = a\cos nt\) and using it; If incorrect \(a\) and/or \(n\) |
| \(v(0.2) = -0.681 \Rightarrow\) velocity is \(0.681ms^{-1}\) upwards | A1ft | If from \(v^2 = n^2(a^2 - x^2)\) the direction must be clearly established |
| [7] |
**Part i**
$2.45e/0.5 = 0.05g$ | M1 | For using $T = \lambda e/L$ and resolving forces vertically; accept use of 0.1 to show both sides equal to 0.49
$(c = 0.1)$ | A1 |
Distance from $O$ is $0.5 + 0.1 = 0.6m$ | A1 |
| [3] |
**Part ii**
$mg - T = m\ddot{x}$ | M1 | For using Newton's second law with 3 terms
$0.05g - 2.45(0.1 + x)/0.5 = 0.05\ddot{x}$ | A1 |
$\ddot{x} = -98x$ | A1 |
| [3] |
**Part iii**
$a = 0.075$ | B1 |
$n = 7\sqrt{2}$ oe | B1 | accept 9.90
$x = 0.075\cos(7\sqrt{2} t)$ | M1 | For using $x = a\cos nt$ oe
$x(0.2) = -0.0298$ | A1 |
$v = -0.075(7\sqrt{2})\sin(7\sqrt{2} t)$ | M1 | For differentiating $x = a\cos nt$ and using it; If incorrect $a$ and/or $n$
$v(0.2) = -0.681 \Rightarrow$ velocity is $0.681ms^{-1}$ upwards | A1ft | If from $v^2 = n^2(a^2 - x^2)$ the direction must be clearly established
| [7] |5 A particle $P$ of mass 0.05 kg is suspended from a fixed point $O$ by a light elastic string of natural length 0.5 m and modulus of elasticity 2.45 N .\\
(i) Show that the equilibrium position of $P$ is 0.6 m below $O$.\\
$P$ is held at rest at a point 0.675 m vertically below $O$ and then released. At time $t \mathrm {~s}$ after $P$ is released, its downward displacement from the equilibrium position is $x \mathrm {~m}$.\\
(ii) Show that $\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } = - 98 x$.\\
(iii) Find the value of $x$ and the magnitude and direction of the velocity of $P$ when $t = 0.2$.
\hfill \mbox{\textit{OCR M3 2011 Q5 [13]}}