| Exam Board | OCR MEI |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2012 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Prove SHM and find period: given force or equation of motion directly |
| Difficulty | Challenging +1.2 This is a standard M3/Further Mechanics SHM question requiring energy conservation, Newton's second law to derive the SHM equation, and application of standard SHM formulas. While it involves multiple parts and requires careful setup (finding natural length via energy, deriving the differential equation, transforming to standard form), each step follows a well-established procedure taught in M3. The derivation of the SHM equation and subsequent calculations are routine for students at this level, though the multi-step nature and need to integrate different mechanics concepts (energy, forces, SHM) makes it moderately above average difficulty. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x6.02d Mechanical energy: KE and PE concepts6.02g Hooke's law: T = k*x or T = lambda*x/l6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| EE is \(\frac{1}{2}(300)x^2\) | B1 | Or \(\frac{1}{2}(300)(40-l_0)^2\); \(x\) is the maximum extension |
| Change in PE is \(75\times9.8\times40\) | B1 | |
| \(\frac{1}{2}(300)x^2 = 75\times9.8\times40\) | M1 | Equation involving EE and PE |
| \(x = 14\); Natural length is 26 m | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(75\times9.8 - 300x = 75\ddot{x}\) | M1, A1 | Equation of motion (three terms) |
| \(\ddot{x} + 4x = 9.8\) | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\ddot{x} = -4(x-2.45)\) | M1 | Or \(\ddot{y}+4(y+c)=9.8\); Condone sign errors for M1 |
| \(c = 2.45\) | A1 | \(c=2.45\) implies M1A1; \(c=-2.45\) implies M1A0 |
| Maximum value of \(y\) is \(14-2.45=11.55\) | B1 | FT is \(37.55-l_0\) (must be positive) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Maximum speed is \(A\omega\) | M1 | |
| \(= 11.55\times2 = 23.1\ \text{ms}^{-1}\) | A1 | FT max value of \(y\) in (iii) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Maximum acceleration is \(A\omega^2\) | Give M1 if M0 for maximum speed | |
| \(= 11.55\times2^2 = 46.2\ \text{ms}^{-2}\) | A1 | Allow \(-46.2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Before rope is stretched, \(26 = \frac{1}{2}\times9.8\times t_1^2\) | B1 | FT from wrong \(l_0\) |
| \(t_1 = 2.304\) | ||
| When rope is stretched, \(y = 11.55\cos 2t\) | M1, A1 | For \(A\sin\omega t\) or \(A\cos\omega t\); For \(11.55\sin 2t\) or \(11.55\cos 2t\); \((t=0\) at lowest point); FT \(A\), \(\omega\) used in (iv) |
| When \(y=-2.45\), \(t_2 = (\pm)0.892\) | M1 | Fully correct strategy for finding \(t_2\) |
| Time to fall \((t_1+t_2)\) is 3.20 s (3 sf) | A1 | CAO |
# Question 3:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| EE is $\frac{1}{2}(300)x^2$ | B1 | Or $\frac{1}{2}(300)(40-l_0)^2$; $x$ is the maximum extension |
| Change in PE is $75\times9.8\times40$ | B1 | |
| $\frac{1}{2}(300)x^2 = 75\times9.8\times40$ | M1 | Equation involving EE and PE |
| $x = 14$; Natural length is 26 m | A1 | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $75\times9.8 - 300x = 75\ddot{x}$ | M1, A1 | Equation of motion (three terms) |
| $\ddot{x} + 4x = 9.8$ | E1 | |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\ddot{x} = -4(x-2.45)$ | M1 | Or $\ddot{y}+4(y+c)=9.8$; Condone sign errors for M1 |
| $c = 2.45$ | A1 | $c=2.45$ implies M1A1; $c=-2.45$ implies M1A0 |
| Maximum value of $y$ is $14-2.45=11.55$ | B1 | FT is $37.55-l_0$ (must be positive) |
## Part (iv)
**(A)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Maximum speed is $A\omega$ | M1 | |
| $= 11.55\times2 = 23.1\ \text{ms}^{-1}$ | A1 | FT max value of $y$ in (iii) |
**(B)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Maximum acceleration is $A\omega^2$ | | Give M1 if M0 for maximum speed |
| $= 11.55\times2^2 = 46.2\ \text{ms}^{-2}$ | A1 | Allow $-46.2$ |
## Part (v)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Before rope is stretched, $26 = \frac{1}{2}\times9.8\times t_1^2$ | B1 | FT from wrong $l_0$ |
| $t_1 = 2.304$ | | |
| When rope is stretched, $y = 11.55\cos 2t$ | M1, A1 | For $A\sin\omega t$ or $A\cos\omega t$; For $11.55\sin 2t$ or $11.55\cos 2t$; $(t=0$ at lowest point); FT $A$, $\omega$ used in (iv) |
| When $y=-2.45$, $t_2 = (\pm)0.892$ | M1 | Fully correct strategy for finding $t_2$ |
| Time to fall $(t_1+t_2)$ is 3.20 s (3 sf) | A1 | CAO |
---3 A bungee jumper of mass 75 kg is connected to a fixed point A by a light elastic rope with stiffness $300 \mathrm { Nm } ^ { - 1 }$. The jumper starts at rest at A and falls vertically. The lowest point reached by the jumper is 40 m vertically below A. Air resistance may be neglected.
\begin{enumerate}[label=(\roman*)]
\item Find the natural length of the rope.
\item Show that, when the rope is stretched and its extension is $x$ metres, $\ddot { x } + 4 x = 9.8$.
The substitution $y = x - c$, where $c$ is a constant, transforms this equation to $\ddot { y } = - 4 y$.
\item Find $c$, and state the maximum value of $y$.
\item Using standard simple harmonic motion formulae, or otherwise, find\\
(A) the maximum speed of the jumper,\\
(B) the maximum deceleration of the jumper.
\item Find the time taken for the jumper to fall from A to the lowest point.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI M3 2012 Q3 [18]}}