| Exam Board | OCR |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2008 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Vertical SHM with two strings |
| Difficulty | Challenging +1.8 This is a challenging M4/Further Mechanics question requiring energy methods to establish equilibrium, then differentiation to derive SHM. The two-particle pulley system with two springs creates algebraic complexity, and the substitution method to prove SHM from energy is more sophisticated than standard direct force approaches. However, it's a structured multi-part question with clear guidance on method. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x6.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(U = 3mgx + 2mg(3a-x) + \frac{mg}{2a}(x-a)^2 + \frac{2mg}{2a}(2a-x)^2\) | B1B1 | Can be awarded for terms listed separately |
| \(= \frac{mg}{2a}(3x^2 - 8ax + 21a^2)\) | ||
| \(\frac{\mathrm{d}U}{\mathrm{d}x} = 3mg - 2mg + \frac{mg}{a}(x-a) - \frac{2mg}{a}(2a-x)\) | M1 | Obtaining \(\frac{\mathrm{d}U}{\mathrm{d}x}\) |
| \(= \frac{3mgx}{a} - 4mg\) | A1 | (or any multiple of this) |
| When \(x = \frac{4}{3}a\), \(\frac{\mathrm{d}U}{\mathrm{d}x} = 4mg - 4mg = 0\) so this is a position of equilibrium | A1 (ag) | |
| \(\frac{\mathrm{d}^2U}{\mathrm{d}x^2} = \frac{3mg}{a}\) | M1 | |
| \(> 0\), so equilibrium is stable | A1 (ag) | Total: 9 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| KE is \(\frac{1}{2}(3m)v^2 + \frac{1}{2}(2m)v^2\) | M1A1 | |
| Energy equation is \(U + \frac{5}{2}mv^2 = \text{constant}\) | M1 | Differentiating the energy equation (with respect to \(t\) or \(x\)) |
| \(\left(\frac{3mgx}{a} - 4mg\right)\frac{\mathrm{d}x}{\mathrm{d}t} + 5mv\frac{\mathrm{d}v}{\mathrm{d}t} = 0\) | A1 ft | |
| \(\frac{3gx}{a} - 4g + 5\frac{\mathrm{d}^2x}{\mathrm{d}t^2} = 0\) | A1 ft | |
| Putting \(x = \frac{4}{3}a + y\): \(\quad \frac{3gy}{a} + 5\frac{\mathrm{d}^2y}{\mathrm{d}t^2} = 0\) | M1A1 ft | Condone \(\ddot{x}\) instead of \(\ddot{y}\); Award M1 even if KE is missing |
| \(\frac{\mathrm{d}^2y}{\mathrm{d}t^2} = -\frac{3g}{5a}y\) | ||
| Hence motion is SHM with period \(2\pi\sqrt{\frac{5a}{3g}}\) | A1 (ag), A1 | Must have \(\ddot{y} = -\omega^2 y\) or other satisfactory explanation. Total: 9 |
# Question 7:
## Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $U = 3mgx + 2mg(3a-x) + \frac{mg}{2a}(x-a)^2 + \frac{2mg}{2a}(2a-x)^2$ | B1B1 | Can be awarded for terms listed separately |
| $= \frac{mg}{2a}(3x^2 - 8ax + 21a^2)$ | | |
| $\frac{\mathrm{d}U}{\mathrm{d}x} = 3mg - 2mg + \frac{mg}{a}(x-a) - \frac{2mg}{a}(2a-x)$ | M1 | Obtaining $\frac{\mathrm{d}U}{\mathrm{d}x}$ |
| $= \frac{3mgx}{a} - 4mg$ | A1 | (or any multiple of this) |
| When $x = \frac{4}{3}a$, $\frac{\mathrm{d}U}{\mathrm{d}x} = 4mg - 4mg = 0$ so this is a position of equilibrium | A1 (ag) | |
| $\frac{\mathrm{d}^2U}{\mathrm{d}x^2} = \frac{3mg}{a}$ | M1 | |
| $> 0$, so equilibrium is stable | A1 (ag) | **Total: 9** |
## Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| KE is $\frac{1}{2}(3m)v^2 + \frac{1}{2}(2m)v^2$ | M1A1 | |
| Energy equation is $U + \frac{5}{2}mv^2 = \text{constant}$ | M1 | Differentiating the energy equation (with respect to $t$ or $x$) |
| $\left(\frac{3mgx}{a} - 4mg\right)\frac{\mathrm{d}x}{\mathrm{d}t} + 5mv\frac{\mathrm{d}v}{\mathrm{d}t} = 0$ | A1 ft | |
| $\frac{3gx}{a} - 4g + 5\frac{\mathrm{d}^2x}{\mathrm{d}t^2} = 0$ | A1 ft | |
| Putting $x = \frac{4}{3}a + y$: $\quad \frac{3gy}{a} + 5\frac{\mathrm{d}^2y}{\mathrm{d}t^2} = 0$ | M1A1 ft | Condone $\ddot{x}$ instead of $\ddot{y}$; Award M1 even if KE is missing |
| $\frac{\mathrm{d}^2y}{\mathrm{d}t^2} = -\frac{3g}{5a}y$ | | |
| Hence motion is SHM with period $2\pi\sqrt{\frac{5a}{3g}}$ | A1 (ag), A1 | Must have $\ddot{y} = -\omega^2 y$ or other satisfactory explanation. **Total: 9** |7\\
\includegraphics[max width=\textwidth, alt={}, center]{a9e010ce-c3a8-4f95-a154-fd16ef3e5e5b-4_622_767_269_689}
Particles $P$ and $Q$, with masses $3 m$ and $2 m$ respectively, are connected by a light inextensible string passing over a smooth light pulley. The particle $P$ is connected to the floor by a light spring $S _ { 1 }$ with natural length $a$ and modulus of elasticity mg . The particle $Q$ is connected to the floor by a light spring $S _ { 2 }$ with natural length $a$ and modulus of elasticity $2 m g$. The sections of the string not in contact with the pulley, and the two springs, are vertical. Air resistance may be neglected. The particles $P$ and $Q$ move vertically and the string remains taut; when the length of $S _ { 1 }$ is $x$, the length of $S _ { 2 }$ is ( $3 a - x$ ) (see diagram).\\
(i) Find the total potential energy of the system (taking the floor as the reference level for gravitational potential energy). Hence show that $x = \frac { 4 } { 3 } a$ is a position of stable equilibrium.\\
(ii) By differentiating the energy equation, and substituting $x = \frac { 4 } { 3 } a + y$, show that the motion is simple harmonic, and find the period.
\hfill \mbox{\textit{OCR M4 2008 Q7 [18]}}