OCR MEI M3 2008 June — Question 3 18 marks

Exam BoardOCR MEI
ModuleM3 (Mechanics 3)
Year2008
SessionJune
Marks18
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSimple Harmonic Motion
TypeTwo springs/strings system equilibrium
DifficultyStandard +0.8 This is a substantial M3/Further Mechanics SHM question requiring multiple steps: finding tensions using Hooke's law, deriving the SHM equation from forces, identifying equilibrium position, calculating period from ω, determining amplitude from initial conditions, and solving for specific time using SHM equations. While methodical, it demands careful bookkeeping of two strings, correct force analysis including weight, and fluency with SHM formulas—significantly above average difficulty but follows standard M3 patterns.
Spec4.10f Simple harmonic motion: x'' = -omega^2 x6.02g Hooke's law: T = k*x or T = lambda*x/l6.02i Conservation of energy: mechanical energy principle
3 A small block B has mass 2.5 kg . A light elastic string connects B to a fixed point P , and a second light elastic string connects \(B\) to a fixed point \(Q\), which is 6.5 m vertically below \(P\). The string PB has natural length 3.2 m and stiffness \(35 \mathrm { Nm } ^ { - 1 }\); the string BQ has natural length 1.8 m and stiffness \(5 \mathrm { Nm } ^ { - 1 }\). The block B is released from rest in the position 4.4 m vertically below P . You are given that B performs simple harmonic motion along part of the line PQ, and that both strings remain taut throughout the motion. Air resistance may be neglected. At time \(t\) seconds after release, the length of the string PB is \(x\) metres (see Fig. 3). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{2a4afead-e772-4d86-bc8d-86ffa5bca507-3_775_345_772_900} \captionsetup{labelformat=empty} \caption{Fig. 3}
\end{figure}
  1. Find, in terms of \(x\), the tension in the string PB and the tension in the string BQ .
  2. Show that \(\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } = 64 - 16 x\).
  3. Find the value of \(x\) when B is at the centre of oscillation.
  4. Find the period of oscillation.
  5. Write down the amplitude of the motion and find the maximum speed of B.
  6. Find the time after release when \(B\) is first moving downwards with speed \(0.9 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
Question 3:
Part (i):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(T_{PB} = 35(x - 3.2)\ [= 35x - 112]\)B1
\(T_{BQ} = 5(5.6 - x - 1.8)\)M1 Finding extension of BQ
\(= 5(4.7 - x)\ [= 23.5 - 5x]\)A1 3 marks total
Part (ii):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(T_{BQ} + mg - T_{PB} = m\frac{d^2x}{dt^2}\)M1 Equation of motion (condone one missing force)
\(5(4.7-x) + 2.5\times9.8 - 35(x-3.2) = 2.5\frac{d^2x}{dt^2}\)A2 Give A1 for three terms correct
\(160 - 40x = 2.5\frac{d^2x}{dt^2}\)
\(\frac{d^2x}{dt^2} = 64 - 16x\)E1 4 marks total
Part (iii):
AnswerMarks Guidance
Answer/WorkingMark Guidance
At the centre, \(\frac{d^2x}{dt^2} = 0\)M1
\(x = 4\)A1 2 marks total
Part (iv):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\omega^2 = 16\)M1 Seen or implied *(allow M1 for \(\omega = 16\))*
Period is \(\frac{2\pi}{\sqrt{16}} = \frac{1}{2}\pi = 1.57\) sA1 Accept \(\frac{1}{2}\pi\); 2 marks total
Part (v):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Amplitude \(A = 4.4 - 4 = 0.4\) mB1 ft ft is \(
Maximum speed is \(A\omega = 0.4 \times 4 = 1.6\ \text{ms}^{-1}\)M1, A1 cao 3 marks total
Part (vi):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(x = 4 + 0.4\cos 4t\)M1 For \(v = C\sin\omega t\) or \(C\cos\omega t\)
\(v = (-)1.6\sin 4t\)A1 *This M1A1 can be earned in (v)*
When \(v = 0.9\), \(\sin 4t = -\frac{0.9}{1.6}\)
\(4t = \pi + 0.5974\)M1 Fully correct method for finding required time; e.g. \(\frac{1}{4}\arcsin\frac{0.9}{1.6} + \frac{1}{2}\text{period}\)
Time is \(0.935\) sA1 cao 4 marks total
*OR* \(0.9^2 = 16(0.4^2 - y^2)\), \(y = -0.3307\)M1 Using \(v^2 = \omega^2(A^2 - y^2)\) *and* \(y = A\cos\omega t\) or \(A\sin\omega t\)
\(y = 0.4\cos 4t\), \(\cos 4t = -\frac{0.3307}{0.4}\)A1 For \(y = (\pm)0.331\) *and* \(y = 0.4\cos 4t\)
\(4t = \pi + 0.5974\), Time is \(0.935\) sM1, A1 cao 
# Question 3:

## Part (i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $T_{PB} = 35(x - 3.2)\ [= 35x - 112]$ | B1 | |
| $T_{BQ} = 5(5.6 - x - 1.8)$ | M1 | Finding extension of BQ |
| $= 5(4.7 - x)\ [= 23.5 - 5x]$ | A1 | **3 marks total** |

## Part (ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $T_{BQ} + mg - T_{PB} = m\frac{d^2x}{dt^2}$ | M1 | Equation of motion (condone one missing force) |
| $5(4.7-x) + 2.5\times9.8 - 35(x-3.2) = 2.5\frac{d^2x}{dt^2}$ | A2 | Give A1 for three terms correct |
| $160 - 40x = 2.5\frac{d^2x}{dt^2}$ | | |
| $\frac{d^2x}{dt^2} = 64 - 16x$ | E1 | **4 marks total** |

## Part (iii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| At the centre, $\frac{d^2x}{dt^2} = 0$ | M1 | |
| $x = 4$ | A1 | **2 marks total** |

## Part (iv):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\omega^2 = 16$ | M1 | Seen or implied *(allow M1 for $\omega = 16$)* |
| Period is $\frac{2\pi}{\sqrt{16}} = \frac{1}{2}\pi = 1.57$ s | A1 | Accept $\frac{1}{2}\pi$; **2 marks total** |

## Part (v):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Amplitude $A = 4.4 - 4 = 0.4$ m | B1 ft | ft is $|4.4 - \text{(iii)}|$ |
| Maximum speed is $A\omega = 0.4 \times 4 = 1.6\ \text{ms}^{-1}$ | M1, A1 cao | **3 marks total** |

## Part (vi):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = 4 + 0.4\cos 4t$ | M1 | For $v = C\sin\omega t$ or $C\cos\omega t$ |
| $v = (-)1.6\sin 4t$ | A1 | *This M1A1 can be earned in (v)* |
| When $v = 0.9$, $\sin 4t = -\frac{0.9}{1.6}$ | | |
| $4t = \pi + 0.5974$ | M1 | Fully correct method for finding required time; e.g. $\frac{1}{4}\arcsin\frac{0.9}{1.6} + \frac{1}{2}\text{period}$ |
| Time is $0.935$ s | A1 cao | **4 marks total** |
| *OR* $0.9^2 = 16(0.4^2 - y^2)$, $y = -0.3307$ | M1 | Using $v^2 = \omega^2(A^2 - y^2)$ *and* $y = A\cos\omega t$ or $A\sin\omega t$ |
| $y = 0.4\cos 4t$, $\cos 4t = -\frac{0.3307}{0.4}$ | A1 | For $y = (\pm)0.331$ *and* $y = 0.4\cos 4t$ |
| $4t = \pi + 0.5974$, Time is $0.935$ s | M1, A1 cao | |

---
3 A small block B has mass 2.5 kg . A light elastic string connects B to a fixed point P , and a second light elastic string connects $B$ to a fixed point $Q$, which is 6.5 m vertically below $P$.

The string PB has natural length 3.2 m and stiffness $35 \mathrm { Nm } ^ { - 1 }$; the string BQ has natural length 1.8 m and stiffness $5 \mathrm { Nm } ^ { - 1 }$.

The block B is released from rest in the position 4.4 m vertically below P . You are given that B performs simple harmonic motion along part of the line PQ, and that both strings remain taut throughout the motion. Air resistance may be neglected. At time $t$ seconds after release, the length of the string PB is $x$ metres (see Fig. 3).

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{2a4afead-e772-4d86-bc8d-86ffa5bca507-3_775_345_772_900}
\captionsetup{labelformat=empty}
\caption{Fig. 3}
\end{center}
\end{figure}

(i) Find, in terms of $x$, the tension in the string PB and the tension in the string BQ .\\
(ii) Show that $\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } = 64 - 16 x$.\\
(iii) Find the value of $x$ when B is at the centre of oscillation.\\
(iv) Find the period of oscillation.\\
(v) Write down the amplitude of the motion and find the maximum speed of B.\\
(vi) Find the time after release when $B$ is first moving downwards with speed $0.9 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.

\hfill \mbox{\textit{OCR MEI M3 2008 Q3 [18]}}
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