OCR M3 2013 January — Question 7 14 marks

Exam BoardOCR
ModuleM3 (Mechanics 3)
Year2013
SessionJanuary
Marks14
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSimple Harmonic Motion
TypeSmall oscillations: simple pendulum (particle on string)
DifficultyStandard +0.3 This is a standard M3 simple pendulum question requiring energy conservation to find maximum angle, verification of SHM conditions for small oscillations, and application of SHM equations. While multi-part with several techniques, all steps follow routine procedures taught in M3 with no novel insight required—slightly easier than average due to straightforward application of standard methods.
Spec1.05q Trig in context: vectors, kinematics, forces6.05d Variable speed circles: energy methods

7 A simple pendulum consists of a light inextensible string of length 0.8 m and a particle \(P\) of mass \(m \mathrm {~kg}\). The pendulum is hanging vertically at rest from a fixed point \(O\) when \(P\) is given a horizontal velocity of \(0.3 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
  1. Show that, in the subsequent motion, the maximum angle between the string and the downward vertical is 0.107 radians, correct to 3 significant figures.
  2. Show that the motion may be modelled as simple harmonic motion, and find the period of this motion.
  3. Find the time after the start of the motion when the velocity of the particle is first \(- 0.2 \mathrm {~ms} ^ { - 1 }\) and find the angular displacement of \(O P\) from the downward vertical at this time.

Question 7:
Part (i):
AnswerMarks Guidance
AnswerMarks Guidance
Use of energy equationM1 Allow M1 if sign error and/or 9.8 missing and/or missing \(m\) or \(l\)
\(0.5\,m\,(0.3)^2 = m \times 9.8 \times 0.8 \times (1 - \cos\theta)\)A1
\(\theta = 0.107\)A1 AG [3] No errors; 0.107194171
Part (ii):
AnswerMarks Guidance
AnswerMarks Guidance
Use \(F = ma\)M1 Allow M1 if sign error, or 9.8 missing
\(\ddot{\theta} = -12.25\,\theta\)A1 \(m \times 9.8\sin\theta = -m \times 0.8\,\ddot{\theta}\)
Small \(\theta\)B1 Dep on having seen \(\text{acc} = k\sin\theta\) or sight of \(\omega = 3.5\)
Use of \(T = \frac{2\pi}{\omega}\)M1 Rigorous
\(T = 1.80\)A1 [5] Accept \(\frac{4\pi}{7}\) (1.795195)
Question 7:
Part (iii):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Identifying amplitude as \(0.107\)B1 Follow through from (i)
Use of \(\dot{\theta} = 0.107 \times 3.5 \times \cos(3.5t)\)M1 or \(\sin(3.5t + \varepsilon)\), \(\varepsilon \neq 0\)
Use of \(\dot{\theta} = -0.25\)A1 Consistent angle or length
\(t = 0.658\)A1 \((0.6576339)\)
Use of \(\theta = 0.107\sin(3.5t)\)M1 Follow through from velocity equation (matches, ignore sign)
\((\theta =)\ 0.0797\) radsA1 Accept \(5.20°\); \((0.0796678\) or \(0.079576)\)
Total[6] Follow through for \(a\) and \(\omega\); allow sign error
# Question 7:

## Part (i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Use of energy equation | M1 | Allow M1 if sign error and/or 9.8 missing and/or missing $m$ or $l$ |
| $0.5\,m\,(0.3)^2 = m \times 9.8 \times 0.8 \times (1 - \cos\theta)$ | A1 | |
| $\theta = 0.107$ | A1 **AG** **[3]** | No errors; 0.107194171 |

## Part (ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Use $F = ma$ | M1 | Allow M1 if sign error, or 9.8 missing |
| $\ddot{\theta} = -12.25\,\theta$ | A1 | $m \times 9.8\sin\theta = -m \times 0.8\,\ddot{\theta}$ |
| Small $\theta$ | B1 | Dep on having seen $\text{acc} = k\sin\theta$ or sight of $\omega = 3.5$ |
| Use of $T = \frac{2\pi}{\omega}$ | M1 | Rigorous |
| $T = 1.80$ | A1 **[5]** | Accept $\frac{4\pi}{7}$ (1.795195) |

## Question 7:

### Part (iii):

| Answer/Working | Marks | Guidance |
|---|---|---|
| Identifying amplitude as $0.107$ | B1 | Follow through from (i) |
| Use of $\dot{\theta} = 0.107 \times 3.5 \times \cos(3.5t)$ | M1 | or $\sin(3.5t + \varepsilon)$, $\varepsilon \neq 0$ |
| Use of $\dot{\theta} = -0.25$ | A1 | Consistent angle or length |
| $t = 0.658$ | A1 | $(0.6576339)$ |
| Use of $\theta = 0.107\sin(3.5t)$ | M1 | Follow through from velocity equation (matches, ignore sign) |
| $(\theta =)\ 0.0797$ rads | A1 | Accept $5.20°$; $(0.0796678$ or $0.079576)$ |
| **Total** | **[6]** | Follow through for $a$ and $\omega$; allow sign error |
7 A simple pendulum consists of a light inextensible string of length 0.8 m and a particle $P$ of mass $m \mathrm {~kg}$. The pendulum is hanging vertically at rest from a fixed point $O$ when $P$ is given a horizontal velocity of $0.3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(i) Show that, in the subsequent motion, the maximum angle between the string and the downward vertical is 0.107 radians, correct to 3 significant figures.\\
(ii) Show that the motion may be modelled as simple harmonic motion, and find the period of this motion.\\
(iii) Find the time after the start of the motion when the velocity of the particle is first $- 0.2 \mathrm {~ms} ^ { - 1 }$ and find the angular displacement of $O P$ from the downward vertical at this time.

\hfill \mbox{\textit{OCR M3 2013 Q7 [14]}}