## Question 7:

### Part 7a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{4(x-2)}{2} = \frac{2(7-x-3)}{3}$ | M1, A1, A1 | Equate tensions in two strings; must use $k\lambda\frac{x}{l}$; sum of extensions must be 2; correct LHS; correct RHS |
| $6(x-2) = 2(4-x)$ | | |
| $x = 2.5$ (m) $\quad *$ | A1 | Given result from fully correct working |

**ALT:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $(T_A =)\frac{4x}{2} = \frac{2y}{3} (= T_B), \quad x + y = 2$ | M1, A1 | Obtain 2 equations; 2 correct equations |
| $x = \frac{1}{2}$ or $y = 1.5$ | A1 | Correct extension for either string |
| Distance $AO = 2.5$ (m) | A1 | Obtain given result from fully correct working |

### Part 7bi:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $2\ddot{y} = \frac{2(1.5-y)}{3} - \frac{4(y+0.5)}{2}$ OR $2\ddot{y} = \frac{4(-y+0.5)}{2} - \frac{2(1.5+y)}{3}$ | M1, A1 | Equation of motion with two variable tensions; allow $a$ or $\ddot{y}$; correct equation |
| $\ddot{y} = -\frac{4}{3}y = -\omega^2 y \; \therefore \text{SHM}$ | M1, A1 | Rearrange to required form (must now be $\ddot{y}$); correct result from fully correct working **and** concluding statement |

### Part 7bii:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\omega^2 = \frac{4}{3}$ | B1 | Correct $\omega$ or $\omega^2$ |
| $(2v_{\max} = 6 \Rightarrow)\; v_{\max} = 3$ (ms$^{-1}$) | B1 | $v_{\max} = 3$ or $2v_{\max} = 6$ seen explicitly or used |
| $v_{\max} = a\omega = \frac{2a}{\sqrt{3}}$ — Accept $1.2a$ or better | M1 | Use $v_{\max} = a\omega$ with their $\omega$, or use $v^2 = \omega^2(a^2 - x^2)$ with $x=0$ |
| $3 = a\dfrac{2}{\sqrt{3}} \quad a = \dfrac{3\sqrt{3}}{2}$ (m) — Accept 2.6 or better | A1 | |

### Part 7c:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $v^2 = \frac{4}{3}\left(\frac{27}{4} - \left(\frac{3}{2}\right)^2\right)$ | M1 | Use $v^2 = \omega^2(a^2 - x^2)$ with $x = 1.5$ and their $\omega$ and $a$, OR energy equation with correct number of terms |
| $v = \sqrt{6}$ (ms$^{-1}$) — Accept 2.4 or better | A1 | |

### Part 7d:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{3}{2} = \frac{3\sqrt{3}}{2}\sin(\omega t)$ | M1, A1 | Use of $x = a\sin(\omega t)$ with $x = 1.5$ and their $\omega$ and $a$; correct equation, ft their $\omega$ and $a$ |
| $t = 0.53$ (s) or better | A1 | $t = 0.533021\ldots$ |

**ALT:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = a\cos\omega t \Rightarrow 1.5 = \frac{3\sqrt{3}}{2}\cos\left(\frac{2}{\sqrt{3}}\right)t$ | M1 | Complete method using cosine |
| time $= \frac{\pi}{2} \times \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2}\cos^{-1}\!\left(\frac{1}{\sqrt{3}}\right)$ | A1ft | Correct equation(s), ft their $\omega$ and $a$ |
| $t = 0.53$ or better | A1 | |

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