| Exam Board | OCR |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2017 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Small oscillations: rigid body compound pendulum |
| Difficulty | Challenging +1.2 This is a standard compound pendulum problem requiring application of the parallel axis theorem and calculus optimization. Part (i) involves straightforward substitution into a given formula with moment of inertia calculation, while part (ii) requires differentiation to minimize—both are well-practiced techniques in M4, though the algebraic manipulation and optimization step elevate it slightly above average difficulty. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force4.10f Simple harmonic motion: x'' = -omega^2 x6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(I = \frac{1}{3}(2M)(2a)^2 + 2M(ka)^2\) | B1 B1 | B1 for each term; SC B1 for incorrect masses in both expressions |
| \(T = 2\pi\sqrt{\frac{8Ma^2 + 6Mk^2a^2}{3(2Mg)(ka)}}\) | M1 | Using \(T = 2\pi\sqrt{\frac{I}{mgh}}\) with correct mass |
| \(T = 2\pi\sqrt{\frac{a}{3g}\left(\frac{4+3k^2}{k}\right)}\) | A1 [4] | AG www |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{d}{dk}\left(\frac{4+3k^2}{3k}\right) = 0\) | M1 | Attempt to differentiate \(f(k)\) and set equal to zero; Or differentiates \(T\) wrt \(k\) |
| \(\frac{3k(6k)-(4+3k^2)(3)}{(3k)^2} = 0 \Rightarrow k^2 = \ldots\) | M1 | Correct application of quotient rule (or other correct method) and get to \(k^2 = \ldots\) or \(k = \ldots\); eg \(\frac{d}{dk}\left(\frac{4}{3k}+k\right)=0\) |
| \(k^2 = \frac{4}{3}\) | A1 [3] | Cao (oe) |
# Question 1:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $I = \frac{1}{3}(2M)(2a)^2 + 2M(ka)^2$ | B1 B1 | B1 for each term; SC B1 for incorrect masses in both expressions |
| $T = 2\pi\sqrt{\frac{8Ma^2 + 6Mk^2a^2}{3(2Mg)(ka)}}$ | M1 | Using $T = 2\pi\sqrt{\frac{I}{mgh}}$ with correct mass |
| $T = 2\pi\sqrt{\frac{a}{3g}\left(\frac{4+3k^2}{k}\right)}$ | A1 **[4]** | AG www |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{d}{dk}\left(\frac{4+3k^2}{3k}\right) = 0$ | M1 | Attempt to differentiate $f(k)$ and set equal to zero; Or differentiates $T$ wrt $k$ |
| $\frac{3k(6k)-(4+3k^2)(3)}{(3k)^2} = 0 \Rightarrow k^2 = \ldots$ | M1 | Correct application of quotient rule (or other correct method) and get to $k^2 = \ldots$ or $k = \ldots$; eg $\frac{d}{dk}\left(\frac{4}{3k}+k\right)=0$ |
| $k^2 = \frac{4}{3}$ | A1 **[3]** | Cao (oe) |
---1 A uniform rod with centre $C$ has mass $2 M$ and length 4a. The rod is free to rotate in a vertical plane about a smooth fixed horizontal axis passing through a point $A$ on the rod, where $A C = k a$ and $0 < k < 2$. The rod is making small oscillations about the equilibrium position with period $T$.\\
(i) Show that $T = 2 \pi \sqrt { \frac { a } { 3 g } \left( \frac { 4 + 3 k ^ { 2 } } { k } \right) }$. (You may assume the standard formula $T = 2 \pi \sqrt { \frac { I } { m g h } }$ for the period of small oscillations of a compound pendulum.)\\
(ii) Hence find the value of $k ^ { 2 }$ for which the period of oscillations is least.
\hfill \mbox{\textit{OCR M4 2017 Q1 [7]}}