**(a)**

$T = \frac{\lambda x}{l} = \frac{\lambda \times 0.5l}{l}$ | M1A1 |

$\lambda = 2mg$ * | A1 | (3)

**(b)**

$mg - T = m\ddot{x}$ | M1 |

$mg - \frac{2mg(0.5l + x)}{l} = m\ddot{x}$ | DM1A1A1 |

$\ddot{x} = -\frac{2gx}{l}$ | A1 |

$\therefore \text{SHM}$ | A1cso(B1 on e-pen) | (6)

**(c)**

$a = 0.3l$ |  |

$|\ddot{x}|_{\max} = 2g \times \frac{0.3l}{l} = 0.6g \left(= 5.88 \text{ or } 5.9 \text{ m s}^{-2}\right)$ | M1A1ft | (2)

**(d)**

$x = a \cos \omega t = 0.3l \cos \sqrt{\frac{2g}{l}}t$ | M1 |

Time C to D: $0.15 = 0.3\cos\sqrt{\frac{2g}{l}}t$ | M1 |

$t = \sqrt{\frac{l}{2g}}\cos^{-1}0.5$ |  |

Time to E: $t' = \text{half period} = \pi\sqrt{\frac{l}{2g}}$ | B1 |

Time D to E: $\left(\pi - \cos^{-1}0.5\right)\sqrt{\frac{l}{2g}} = \frac{2\pi}{3}\sqrt{\frac{l}{2g}}$ | M1A1 | (4) [15] |

**Notes for Question 7(a):**

M1 | for using Hooke's Law |

A1 | for a correct equation |

A1 | for solving to get $\lambda = 2mg$ * |

**(b)**

M1 | for using NL2. Weight and tension must be seen. Acceleration can be $a$ here, but must be an equation at a general position |

M1 dep | for using Hooke's Law for the tension. Acceleration can be $a$ |

A1 A1 | for a fully correct equation. inc acceleration as $\ddot{x}$ (-1 ce) |

A1 | for simplifying to $\ddot{x} = -\frac{2gx}{l}$ oe |

A1 cso | for the conclusion |

**(c)**

M1 | for using $|\ddot{x}|_{\max} = \omega^2 a$ with their $\omega$ and $a = 0.3l$. $\omega$ must be dimensionally correct |

A1 ft | for obtaining the max magnitude of the accel, accept 0.6g, 5.9 or 5.88 only. ft their $\omega$ |

**(d)**

M1 | for using $x = a \cos \omega t$ with $x = \pm 0.15l$, $a = 0.3l$ and their $\omega$ to obtain an expression for the time from C to D |

B1 | for time C to E $=$ half period $= \pi\sqrt{\frac{l}{2g}}$ |

M1 | For any correct method for obtaining the time from D to E |

A1 cso | for $\frac{2\pi}{3}\sqrt{\frac{l}{2g}}$ oe inc $0.473\sqrt{l}$ $0.47\sqrt{l}$ |

**ALT for (d)(i):**

M1 | Use $x = a \sin \omega t$ with $x = 0.15l$, $a = 0.3l$ and their $\omega$ to obtain an expression for the time from B to D |

M1, A1 | as above |

Using $x = a \cos \omega t$ with $x = \pm 0.15l$, $a = 0.3l$ and their $\omega$ |

This gives the required time in one step. |

Award M2 A1 for correct substitution |

A1 correct answer |

However do not isw if further work shown. Mark according to mark scheme method and give max M1B1M0a0. |

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## Question 8: [Alternative for Question 7(d)]

By reference circle:

Centre of circle is O
Angle $COD = \theta$, Angle $EOD = \alpha$

$\cos \theta = \frac{0.15l}{0.3l}$, $\theta = \frac{\pi}{3}$ | M1 |

$\alpha = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$ | B1 |

$\omega = \sqrt{\frac{2g}{l}}$ |  |

time $= \frac{\alpha}{\omega} = \frac{2\pi/3}{\sqrt{\frac{2g}{l}}} = \frac{2\pi}{3}\sqrt{\frac{l}{2g}}$ | M1A1 |