| Exam Board | OCR MEI |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2006 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Vertical SHM with two strings |
| Difficulty | Challenging +1.2 This is a standard Further Maths Mechanics 3 vertical SHM question with two elastic strings. While it requires multiple techniques (equilibrium verification, tension formulas, SHM equation derivation, and energy/kinematics), each part follows predictable patterns taught in M3. The multi-part structure and need to track two strings simultaneously elevates it slightly above average difficulty, but it remains a textbook-style question without requiring novel insight. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x6.02g Hooke's law: T = k*x or T = lambda*x/l6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Tension in AP: \(T_{AP} = \dfrac{1323(AP - 3)}{3}\) | M1 | Using \(T = \frac{\lambda e}{l_0}\) |
| Tension in BP: \(T_{BP} = \dfrac{1323(BP - 4.5)}{4.5}\), where \(BP = 12 - AP\) | A1 | |
| Equilibrium: \(T_{AP} = T_{BP} + 15g\) | M1 | |
| \(\dfrac{1323(AP-3)}{3} = \dfrac{1323(12-AP-4.5)}{4.5} + 147\) | A1 | |
| Solving gives \(AP = 5\) m ✓ | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(AP = 5 + x\), extension in AP \(= (2+x)\) m | M1 | |
| \(T_{AP} = \dfrac{1323(2+x)}{3} = 441(2+x)\) N | A1 | |
| \(BP = 7 - x\), extension in BP \(= (7-x-4.5) = (2.5-x)\) m | M1 | |
| \(T_{BP} = \dfrac{1323(2.5-x)}{4.5} = 294(2.5-x)\) N | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Net upward force \(= T_{AP} - T_{BP} - 15g\) | M1 | |
| \(= 441(2+x) - 294(2.5-x) - 147\) | A1 | |
| \(= 882 + 441x - 735 + 294x - 147 = 735x\) | A1 | |
| \(15\ddot{x} = -735x \Rightarrow \ddot{x} = -49x\) | M1 | SHM condition |
| \(\omega^2 = 49\), \(T = \dfrac{2\pi}{7} \approx 0.898\) s | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Min AP \(= 3.5\) m \(\Rightarrow x = -1.5\) m (amplitude \(= 1.5\) m) | M1 | |
| Max AP \(= 5 + 1.5 = 6.5\) m | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(x = 4.1 - 5 = -0.9\) m | B1 | |
| \(v^2 = \omega^2(a^2 - x^2) = 49(1.5^2 - 0.9^2)\) | M1 | |
| \(v^2 = 49(2.25 - 0.81) = 49 \times 1.44 = 70.56\) | A1 | |
| \(v = 8.4\) ms\(^{-1}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(x = -1.5\) when \(AP = 3.5\): \(x = -1.5\cos(7t)\) | M1 | Starting from minimum |
| When \(AP = 4.5\): \(x = -0.5\), so \(-0.5 = -1.5\cos(7t)\) | M1 | |
| \(\cos(7t) = \dfrac{1}{3}\) | A1 | |
| \(t = \dfrac{1}{7}\arccos\!\left(\dfrac{1}{3}\right) = \dfrac{1.2310}{7} \approx 0.176\) s | A1 |
# Question 3:
## Part (i) - Verify AP = 5 m
| Answer | Mark | Guidance |
|--------|------|----------|
| Tension in AP: $T_{AP} = \dfrac{1323(AP - 3)}{3}$ | M1 | Using $T = \frac{\lambda e}{l_0}$ |
| Tension in BP: $T_{BP} = \dfrac{1323(BP - 4.5)}{4.5}$, where $BP = 12 - AP$ | A1 | |
| Equilibrium: $T_{AP} = T_{BP} + 15g$ | M1 | |
| $\dfrac{1323(AP-3)}{3} = \dfrac{1323(12-AP-4.5)}{4.5} + 147$ | A1 | |
| Solving gives $AP = 5$ m ✓ | A1 | |
## Part (ii) - Tensions
| Answer | Mark | Guidance |
|--------|------|----------|
| $AP = 5 + x$, extension in AP $= (2+x)$ m | M1 | |
| $T_{AP} = \dfrac{1323(2+x)}{3} = 441(2+x)$ N | A1 | |
| $BP = 7 - x$, extension in BP $= (7-x-4.5) = (2.5-x)$ m | M1 | |
| $T_{BP} = \dfrac{1323(2.5-x)}{4.5} = 294(2.5-x)$ N | A1 | |
## Part (iii) - Simple harmonic motion and period
| Answer | Mark | Guidance |
|--------|------|----------|
| Net upward force $= T_{AP} - T_{BP} - 15g$ | M1 | |
| $= 441(2+x) - 294(2.5-x) - 147$ | A1 | |
| $= 882 + 441x - 735 + 294x - 147 = 735x$ | A1 | |
| $15\ddot{x} = -735x \Rightarrow \ddot{x} = -49x$ | M1 | SHM condition |
| $\omega^2 = 49$, $T = \dfrac{2\pi}{7} \approx 0.898$ s | A1 | |
## Part (iv) - Maximum length of AP
| Answer | Mark | Guidance |
|--------|------|----------|
| Min AP $= 3.5$ m $\Rightarrow x = -1.5$ m (amplitude $= 1.5$ m) | M1 | |
| Max AP $= 5 + 1.5 = 6.5$ m | A1 | |
## Part (v) - Speed when AP = 4.1 m
| Answer | Mark | Guidance |
|--------|------|----------|
| $x = 4.1 - 5 = -0.9$ m | B1 | |
| $v^2 = \omega^2(a^2 - x^2) = 49(1.5^2 - 0.9^2)$ | M1 | |
| $v^2 = 49(2.25 - 0.81) = 49 \times 1.44 = 70.56$ | A1 | |
| $v = 8.4$ ms$^{-1}$ | A1 | |
## Part (vi) - Time for AP to increase from 3.5 m to 4.5 m
| Answer | Mark | Guidance |
|--------|------|----------|
| $x = -1.5$ when $AP = 3.5$: $x = -1.5\cos(7t)$ | M1 | Starting from minimum |
| When $AP = 4.5$: $x = -0.5$, so $-0.5 = -1.5\cos(7t)$ | M1 | |
| $\cos(7t) = \dfrac{1}{3}$ | A1 | |
| $t = \dfrac{1}{7}\arccos\!\left(\dfrac{1}{3}\right) = \dfrac{1.2310}{7} \approx 0.176$ s | A1 | |
---3 A fixed point A is 12 m vertically above a fixed point B. A light elastic string, with natural length 3 m and modulus of elasticity 1323 N , has one end attached to A and the other end attached to a particle P of mass 15 kg . Another light elastic string, with natural length 4.5 m and modulus of elasticity 1323 N , has one end attached to B and the other end attached to P .\\
(i) Verify that, in the equilibrium position, $\mathrm { AP } = 5 \mathrm {~m}$.
The particle P now moves vertically, with both strings AP and BP remaining taut throughout the motion. The displacement of P above the equilibrium position is denoted by $x \mathrm {~m}$ (see Fig. 3).
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{5bb02383-91c0-4454-aaea-0bd6af6ba325-4_405_360_751_849}
\captionsetup{labelformat=empty}
\caption{Fig. 3}
\end{center}
\end{figure}
(ii) Show that the tension in the string AP is $441 ( 2 - x ) \mathrm { N }$ and find the tension in the string BP .\\
(iii) Show that the motion of P is simple harmonic, and state the period.
The minimum length of AP during the motion is 3.5 m .\\
(iv) Find the maximum length of AP .\\
(v) Find the speed of P when $\mathrm { AP } = 4.1 \mathrm {~m}$.\\
(vi) Find the time taken for AP to increase from 3.5 m to 4.5 m .
\hfill \mbox{\textit{OCR MEI M3 2006 Q3 [18]}}