# Question 7:

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $T_A - T_B = m\ddot{x}$ | M1 | Equation of motion in a general position, allow $a$ for acceleration, correct no. of terms, condone sign errors |
| $\frac{2mg}{l}\left(\frac{2l}{3}-x\right) - \frac{mg}{l}\left(\frac{4l}{3}+x\right) = m\ddot{x}$ or $\frac{mg}{l}\left(\frac{4l}{3}-x\right) - \frac{2mg}{l}\left(\frac{2l}{3}+x\right) = m\ddot{x}$ | dM1 A1 | Use Hooke's Law to sub for the two tensions; extensions must be different and of the form $(d \pm x)$ where $d$ is a multiple of $l$. Correct unsimplified equation |
| $-\frac{3g}{l}x = \ddot{x}$, so SHM | A1 | Correct equation using $\ddot{x}$ for acceleration |
| $T = \frac{2\pi}{\sqrt{\frac{3g}{l}}} = 2\pi\sqrt{\frac{l}{3g}}$ * | M1 A1* | Use of $\frac{2\pi}{\omega}$, their $\omega$ from equation of motion which must be in terms of $x$. Given answer correctly obtained with conclusion and correct period |

## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2}l \times \sqrt{\frac{3g}{l}}$ or $\frac{1}{2}\sqrt{3gl}$ or $\sqrt{\frac{3gl}{4}}$ oe | B1 | Speed at $O$ so must be positive. Unsimplified, ignore errors from subsequent 'simplifying' of surds |

## Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{3g}{2}$ or $1.5g$ | B1 | Max acceleration so must be positive |

## Part (d):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = a\cos\omega t \Rightarrow v = -a\omega\sin\omega t$ | M1 | Use of $x = a\cos\omega t$ to obtain $v = -a\omega\sin\omega t$. Substitution for $a$ and $\omega$ not required |
| $-\frac{3}{4}\sqrt{gl} = -a\omega\sin\omega t$ to find $t$ | M1 A1 | Use $v = -a\omega\sin\omega t$ with $a = \frac{l}{2}$ and $\omega = \sqrt{\frac{3g}{l}}$ to obtain equation in $t$ only. Correct equation in $t$ only |
| Solve for $t$ | M1 | Solve to find the required time $t$ |
| $t = \frac{\pi}{3}\sqrt{\frac{l}{3g}}$ oe | A1 | Cao for required time |

### Alternative Methods for Part (d):

**ALT 1:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = a\sin\omega t \Rightarrow v = a\omega\cos\omega t$ | M1 | Substitution for $a$ and $\omega$ not required |
| $\frac{3}{4}\sqrt{gl} = a\omega\cos\omega t$ with $a=\frac{l}{2}$, $\omega=\sqrt{\frac{3g}{l}}$ | M1 A1 | Correct equation in $t$ only |
| Solve for $t$, subtract from $\frac{1}{4}$ period: $t = \frac{\pi}{6}\sqrt{\frac{l}{3g}} \Rightarrow$ required time $= \frac{1}{4}\left(2\pi\sqrt{\frac{l}{3g}}\right) - \frac{\pi}{6}\sqrt{\frac{l}{3g}} = \frac{\pi}{3}\sqrt{\frac{l}{3g}}$ | M1 | Solve to find $t$ then subtract from $\frac{1}{4}$ period |
| $t = \frac{\pi}{3}\sqrt{\frac{l}{3g}}$ oe | A1 | Cao for required time |

**ALT 2:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of $x = a\cos\omega t$ or $x = a\sin\omega t$ | M1 | Substitution for $a$ and $\omega$ not required |
| $v^2 = \omega^2(a^2-x^2)$ with $a=\frac{l}{2}$, $\omega=\sqrt{\frac{3g}{l}}$: $\left(-\frac{3}{4}\sqrt{gl}\right)^2 = \omega^2(a^2-x^2)$ | M1 A1 | Correct equation in $x$ only. Solution leads onto first M mark in (d) |
| $\frac{l}{4} = \frac{l}{2}\cos\left(\sqrt{\frac{3g}{l}}\,t\right)$ or quarter period with sin method | M1 | Solves for $t$ and completes method to find required time |
| $t = \frac{\pi}{3}\sqrt{\frac{l}{3g}}$ oe | A1 | Cao for required time |

### Special Case where $a = \frac{l}{2}$ is clearly stated as amplitude and consistently used in (b), (c) & (d):

| Part | Mark | Answer |
|---|---|---|
| (b) | B1 | $\frac{1}{2}\sqrt{\frac{3g}{l}}$ |
| (c) | B1 | $\frac{3g}{2l}$ |
| (d) | Maximum M1 M1 A0 M0 A0 | — |