Modeling context with interpretation

1. The differential equation

$$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 6 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 13 x = 8 \mathrm { e } ^ { - 3 t } \quad t \geqslant 0$$ describes the motion of a particle along the \(x\)-axis.

1. Determine the general solution of this differential equation. Given that the motion of the particle satisfies \(x = \frac { 1 } { 2 }\) and \(\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { 1 } { 2 }\) when \(t = 0\)
2. determine the particular solution for the motion of the particle. On the graph of the particular solution found in part (b), the first turning point for \(t > 0\) occurs at \(x = a\).
3. Determine, to 3 significant figures, the value of \(a\).
   [0pt] [Solutions relying entirely on calculator technology are not acceptable.]

7. (a) Find the general solution of the differential equation $$2 \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 5 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 2 x = 2 t + 9$$ (b) Find the particular solution of this differential equation for which \(x = 3\) and \(\frac { \mathrm { d } x } { \mathrm {~d} t } = - 1\) when \(t = 0\). The particular solution in part (b) is used to model the motion of a particle \(P\) on the \(x\)-axis. At time \(t\) seconds \(( t \geq 0 ) , P\) is \(x\) metres from the origin \(O\).
(c) Show that the minimum distance between \(O\) and \(P\) is \(\frac { 1 } { 2 } ( 5 + \ln 2 ) \mathrm { m }\) and justify that the distance is a minimum.
(4)(Total 14 marks)

8. $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 5 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 6 x = 2 \mathrm { e } ^ { - t }$$ Given that \(x = 0\) and \(\frac { \mathrm { d } x } { \mathrm {~d} t } = 2\) at \(t = 0\),

1. find \(x\) in terms of \(t\). The solution to part (a) is used to represent the motion of a particle \(P\) on the \(x\)-axis. At time \(t\) seconds, where \(t > 0 , P\) is \(x\) metres from the origin \(O\).
2. Show that the maximum distance between \(O\) and \(P\) is \(\frac { 2 \sqrt { } 3 } { 9 } \mathrm {~m}\) and justify that this
   distance is a maximum.

8 The value of the assets of a large commercial organisation at time \(t\), measured in years, is \(\\) \left( 10 ^ { 8 } y + 10 ^ { 9 } \right)\(. The variables \)y\( and \)t$ are related by the differential equation $$\frac { d ^ { 2 } y } { d t ^ { 2 } } + 5 \frac { d y } { d t } + 6 y = 15 \cos 3 t - 3 \sin 3 t$$ Find \(y\) in terms of \(t\), given that \(y = 3\) and \(\frac { \mathrm { d } y } { \mathrm {~d} t } = - 2\) when \(t = 0\). Show that, for large values of \(t\), the value of the assets is less than \(\\) 9.5 \times 10 ^ { 8 }$ for about a third of the time.

8 A biologist is studying the effect of pesticides on crops. On a certain farm pesticide is regularly applied to a particular crop which grows in soil. Over time, pesticide is transferred between the crop and the soil at a rate which depends on the amount of pesticide in both the crop and the soil. The amount of pesticide in the crop after \(t\) days is \(x\) grams. The amount of pesticide in the soil after \(t\) days is \(y\) grams. Initially, when \(t = 0\), there is no pesticide in either the crop or the soil. At first it is assumed that no pesticide is lost from the system. The biologist further assumes that pesticide is added to the crop at a constant rate of \(k\) grams per day, where \(k > 6\). After collecting some initial data, the biologist suggests that for \(t \geqslant 0\), this situation can be modelled by the following pair of first order linear differential equations. \(\frac { d x } { d t } = - 2 x + 78 y + k\) \(\frac { d y } { d t } = 2 x - 78 y\)

1. 1. Show that \(\frac { d ^ { 2 } x } { d t ^ { 2 } } + 80 \frac { d x } { d t } = 78 \mathrm { k }\).
   2. Determine the particular solution for \(x\) in terms of \(k\) and \(t\). If more than 250 grams of pesticide is found in the crop, then it will fail food safety standards.
   3. The crop is tested 50 days after the pesticide is first added to it. Explain why, according to this model, the crop will fail food safety standards as a result of this test. Further data collection suggests that some pesticide decays in the soil and so is lost from the system. The model is refined in light of this data. The particular solution for \(x\) for this refined model is \(\mathrm { x } = \mathrm { k } \left( 20 - \mathrm { e } ^ { - 41 \mathrm { t } } \left( 20 \cosh ( \sqrt { 1677 } \mathrm { t } ) + \frac { 819 } { \sqrt { 1677 } } \sinh ( \sqrt { 1677 } \mathrm { t } ) \right) \right.\).
2. Given now that \(k < 12\), determine whether the crop will fail food safety standards in the long run according to this refined model. In the refined model, it is still assumed that pesticide is added to the crop at a constant rate.
3. Suggest a reason why it might be more realistic to model the addition of pesticide as not being at a constant rate.

11 The displacement of a door from its equilibrium (closed) position is measured by the angle, \(\theta\) radians, which the door makes with its closed position. The door can swing either side of the equilibrium position so that \(\theta\) can take positive and negative values. The door is released from rest from an open position at time \(t = 0\). A proposed differential equation to model the motion of the door for \(t \geqslant 0\) is \(\frac { \mathrm { d } ^ { 2 } \theta } { \mathrm {~d} t ^ { 2 } } + \lambda \frac { \mathrm { d } \theta } { \mathrm { d } t } + 3 \theta = 0\) where \(\lambda\) is a constant and \(\lambda \geqslant 0\).

1. 1. According to the model, for what value of \(\lambda\) will the motion of the door be simple harmonic?
   2. Explain briefly why modelling the motion of the door as simple harmonic is unlikely to be realistic.
2. Find the range of values of \(\lambda\) for which the model predicts that the door will never pass through the equilibrium position.
3. Sketch a possible graph of \(\theta\) against \(t\) when \(\lambda\) lies outside the range found in part (b) but the motion is not simple harmonic.

11 A company is required to weigh any goods before exporting them overseas. When a crate is placed on a set of weighing scales, the mass displayed takes time to settle down to its final value. The company wishes to model the mass, \(m \mathrm {~kg}\), which is displayed \(t\) seconds after a crate X is placed on the scales.
For the displayed mass it is assumed that the rate of change of the quantity \(\left( 0.5 \frac { \mathrm {~d} m } { \mathrm {~d} t } + m \right)\) with respect to time is proportional to \(( 80 - m )\).

1. Show that \(\frac { \mathrm { d } ^ { 2 } m } { \mathrm {~d} t ^ { 2 } } + 2 \frac { \mathrm {~d} m } { \mathrm {~d} t } + 2 \mathrm {~km} = 160 \mathrm { k }\), where \(k\) is a real constant. It is given that the complementary function for the differential equation in part (i) is \(\mathrm { e } ^ { \lambda t } ( A \cos 2 t + B \sin 2 t )\), where \(A\) and \(B\) are arbitrary constants.
2. Show that \(k = \frac { 5 } { 2 }\) and state the value of the constant \(\lambda\). When X is initially placed on the scales the displayed mass is zero and the rate of increase of the displayed mass is \(160 \mathrm {~kg} \mathrm {~s} ^ { - 1 }\).
3. Find \(m\) in terms of \(t\).
4. Describe the long term behaviour of \(m\).
5. With reference to your answer to part (iv), comment on a limitation of the model.
6. (a) Find the value of \(m\) that corresponds to the stationary point on the curve \(m = \mathrm { f } ( t )\) with the smallest positive value of \(t\).
   (b) Interpret this value of \(m\) in the context of the model.
7. Adapt the differential equation \(\frac { \mathrm { d } ^ { 2 } m } { \mathrm {~d} t ^ { 2 } } + 2 \frac { \mathrm {~d} m } { \mathrm {~d} t } + 5 m = 400\) to model the mass displayed \(t\) seconds after a crate Y , of mass 100 kg , is placed on the scales. \section*{END OF QUESTION PAPER} \section*{Copyright Information:} OCR is committed to seeking permission to reproduce all third-party content that it uses in the assessment materials. OCR has attempted to identify and contact all copyright holders whose work is used in this paper. To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced in the OCR Copyright Acknowledgements booklet. This is produced for each series of examinations and is freely available to download from our public website (\href{http://www.ocr.org.uk}{www.ocr.org.uk}) after the live examination series. If OCR has unwittingly failed to correctly acknowledge or clear any third-party content in this assessment material, OCR will be happy to correct its mistake at the earliest possible opportunity. For queries or further information please contact the Copyright Team, First Floor, 9 Hills Road, Cambridge CB2 1GE.
   OCR is part of the Cambridge Assessment Group; Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge.

6 A particle, \(P\), positioned at the origin, \(O\), is projected with a certain velocity along the \(x\)-axis. \(P\) is then acted on by a single force which varies in such a way that \(P\) moves backwards and forwards along the \(x\)-axis. When the time after projection is \(t\) seconds, the displacement of \(P\) from the origin is \(x \mathrm {~m}\) and its velocity is \(v \mathrm {~ms} ^ { - 1 }\). The motion of \(P\) is modelled using the differential equation \(\ddot { x } + \omega ^ { 2 } x = 0\), where \(\omega\) rads \(^ { - 1 }\) is a positive constant.

1. Write down the general solution of this differential equation. \(D\) is the point where \(x = d\) for some positive constant, \(d\). When \(P\) reaches \(D\) it comes to instantaneous rest.
2. Using the answer to part (a), determine expressions, in terms of \(\omega\), \(d\) and \(t\) only, for the following quantities
   • \(X\)
   • \(v\)
   • Hence show that, according to the model, \(v ^ { 2 } = \omega ^ { 2 } \left( d ^ { 2 } - x ^ { 2 } \right)\).
   The quantity \(z\) is defined by \(z = \frac { 1 } { v }\).
3. Using part (c), determine an expression for \(\mathrm { Z } _ { \mathrm { m } }\), the mean value of z with respect to the displacement, as \(P\) moves directly from \(O\) to \(D\). One measure of the validity of the model is consideration of the value of \(\mathrm { z } _ { \mathrm { m } }\). If \(\mathrm { z } _ { \mathrm { m } }\) exceeds 8 then the model is considered to be valid. The value of \(d\) is measured as 0.25 to 2 significant figures. The value of \(\omega\) is measured as \(0.75 \pm 0.02\).
4. Determine what can be inferred about the validity of the model from the given information.
5. Find, according to the model, the least possible value of the velocity with which \(P\) was initially projected. Give your answer to \(\mathbf { 2 }\) significant figures.

4. A particle \(P\) of mass \(m\) is attached to the mid-point of a light elastic string, of natural length \(2 L\) and modulus of elasticity \(2 m k ^ { 2 } L\), where \(k\) is a positive constant. The ends of the string are attached to points \(A\) and \(B\) on a smooth horizontal surface, where \(A B = 3 L\). The particle is released from rest at the point \(C\), where \(A C = 2 L\) and \(A C B\) is a straight line. During the subsequent motion \(P\) experiences air resistance of magnitude \(2 m k v\), where \(v\) is the speed of \(P\). At time \(t , A P = 1.5 L + x\).

1. Show that \(\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 2 k \frac { \mathrm {~d} x } { \mathrm {~d} t } + 4 k ^ { 2 } x = 0\).
2. Find an expression, in terms of \(t , k\) and \(L\), for the distance \(A P\) at time \(t\).

7. A particle of mass \(m\) is attached to one end \(P\) of a light elastic spring \(P Q\), of natural length \(a\) and modulus of elasticity \(m a n ^ { 2 }\). At time \(t = 0\), the particle and the spring are at rest on a smooth horizontal table, with the spring straight but unstretched and uncompressed. The end \(Q\) of the spring is then moved in a straight line, in the direction \(P Q\), with constant acceleration \(f\). At time \(t\), the displacement of the particle in the direction \(P Q\) from its initial position is \(x\) and the length of the spring is \(( a + y )\).

1. Show that \(x + y = \frac { 1 } { 2 } f t ^ { 2 }\).
2. Hence show that $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + n ^ { 2 } x = \frac { 1 } { 2 } n ^ { 2 } f t ^ { 2 }$$ You are given that the general solution of this differential equation is $$x = A \cos n t + B \sin n t + \frac { 1 } { 2 } f t ^ { 2 } - \frac { f } { n ^ { 2 } }$$ where \(A\) and \(B\) are constants.
3. Find the values of \(A\) and \(B\).
4. Find the maximum tension in the spring. END

4. A particle \(P\) of mass \(m\) is suspended from a fixed point by a light elastic spring. The spring has natural length \(a\) and modulus of elasticity \(2 m \omega ^ { 2 } a\), where \(\omega\) is a positive constant. At time \(t = 0\) the particle is projected vertically downwards with speed \(U\) from its equilibrium position. The motion of the particle is resisted by a force of magnitude \(2 m \omega v\), where \(v\) is the speed of the particle. At time \(t\), the displacement of \(P\) downwards from its equilibrium position is \(x\).

1. Show that \(\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 2 \omega \frac { \mathrm {~d} x } { \mathrm {~d} t } + 2 \omega ^ { 2 } x = 0\). Given that the solution of this differential equation is \(x = \mathrm { e } ^ { - \omega t } ( A \cos \omega t + B \sin \omega t )\), where \(A\) and \(B\) are constants,
2. find \(A\) and \(B\).
3. Find an expression for the time at which \(P\) first comes to rest.
   (3)

6. \begin{figure}[h] \end{figure} A particle \(P\) of mass 2 kg is attached to the mid-point of a light elastic spring of natural length 2 m and modulus of elasticity 4 N . One end \(A\) of the elastic spring is attached to a fixed point on a smooth horizontal table. The spring is then stretched until its length is 4 m and its other end \(B\) is held at a point on the table where \(A B = 4 \mathrm {~m}\). At time \(t = 0 , P\) is at rest on the table at the point \(O\) where \(A O = 2 \mathrm {~m}\), as shown in Fig. 3. The end \(B\) is now moved on the table in such a way that \(A O B\) remains a straight line. At time \(t\) seconds, \(A B = \left( 4 + \frac { 1 } { 2 } \sin 4 t \right) \mathrm { m }\) and \(A P = ( 2 + x ) \mathrm { m }\).

1. Show that $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 4 x = \sin 4 t$$
2. Hence find the time when \(P\) first comes to instantaneous rest. END

5. A light elastic string, of natural length \(2 a\) and modulus of elasticity \(m g\), has a particle \(P\) of mass \(m\) attached to its mid-point. One end of the string is attached to a fixed point \(A\) and the other end is attached to a fixed point \(B\) which is at a distance \(4 a\) vertically below \(A\).

1. Show that \(P\) hangs in equilibrium at the point \(E\) where \(A E = \frac { 5 } { 2 } a\). The particle \(P\) is held at a distance \(3 a\) vertically below \(A\) and is released from rest at time \(t = 0\). When the speed of the particle is \(v\), there is a resistance to motion of magnitude \(2 m k v\), where \(k = \sqrt { } \left( \frac { g } { a } \right)\). At time \(t\) the particle is at a distance \(\left( \frac { 5 } { 2 } a + x \right)\) from \(A\).
2. Show that $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 2 k \frac { \mathrm {~d} x } { \mathrm {~d} t } + 2 k ^ { 2 } x = 0$$
3. Hence find \(x\) in terms of \(t\).

6. A small ball is attached to one end of a spring. The ball is modelled as a particle of mass 0.1 kg and the spring is modelled as a light elastic spring \(A B\), of natural length 0.5 m and modulus of elasticity 2.45 N . The particle is attached to the end \(B\) of the spring. Initially, at time \(t = 0\), the end \(A\) is held at rest and the particle hangs at rest in equilibrium below \(A\) at the point \(E\). The end \(A\) then begins to move along the line of the spring in such a way that, at time \(t\) seconds, \(t \leq 1\), the downward displacement of \(A\) from its initial position is 2 sin \(2 t\) metres. At time \(t\) seconds, the extension of the spring is \(x\) metres and the displacement of the particle below \(E\) is \(y\) metres.

1. Show, by referring to a simple diagram, that \(y + 0.2 = x + 2 \sin 2 t\).
2. Hence show that \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + 49 y = 98 \sin 2 t\). Given that \(y = \frac { 98 } { 45 } \sin 2 t\) is a particular integral of this differential equation,
3. find \(y\) in terms of \(t\).
4. Find the time at which the particle first comes to instantaneous rest.

5. A light elastic spring has natural length \(l\) and modulus of elasticity \(m g\). One end of the spring is fixed to a point \(O\) on a rough horizontal table. The other end is attached to a particle \(P\) of mass \(m\) which is at rest on the table with \(O P = l\). At time \(t = 0\) the particle is projected with speed \(\sqrt { } ( g l )\) along the table in the direction \(O P\). At time \(t\) the displacement of \(P\) from its initial position is \(x\) and its speed is \(v\). The motion of \(P\) is subject to air resistance of magnitude \(2 m v \omega\), where \(\omega = \sqrt { \frac { g } { l } }\). The coefficient of friction between \(P\) and the table is 0.5 .

1. Show that, until \(P\) first comes to rest, $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 2 \omega \frac { \mathrm {~d} x } { \mathrm {~d} t } + \omega ^ { 2 } x = - 0.5 g$$
2. Find \(x\) in terms of \(t , l\) and \(\omega\).
3. Hence find, in terms of \(\omega\), the time taken for \(P\) to first come to instantaneous rest.
   (3)

6. A light elastic spring \(A B\) has natural length \(2 a\) and modulus of elasticity \(2 m n ^ { 2 } a\), where \(n\) is a constant. A particle \(P\) of mass \(m\) is attached to the end \(A\) of the spring. At time \(t = 0\), the spring, with \(P\) attached, lies at rest and unstretched on a smooth horizontal plane. The other end \(B\) of the spring is then pulled along the plane in the direction \(A B\) with constant acceleration \(f\). At time \(t\) the extension of the spring is \(x\).

1. Show that $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + n ^ { 2 } x = f .$$
2. Find \(x\) in terms of \(n , f\) and \(t\). Hence find
3. the maximum extension of the spring,
4. the speed of \(P\) when the spring first reaches its maximum extension.
   \section*{June 2009}

7. A particle \(P\) of mass 0.5 kg is attached to the end \(A\) of a light elastic spring \(A B\), of natural length 0.6 m and modulus of elasticity 2.7 N . At time \(t = 0\) the end \(B\) of the spring is held at rest and \(P\) hangs at rest at the point \(C\) which is vertically below \(B\). The end \(B\) is then moved along the line of the spring so that, at time \(t\) seconds, the downwards displacement of \(B\) from its initial position is \(4 \sin 2 t\) metres. At time \(t\) seconds, the extension of the spring is \(x\) metres and the displacement of \(P\) below \(C\) is \(y\) metres.

1. Show that $$y + \frac { 49 } { 45 } = x + 4 \sin 2 t$$
2. Hence show that $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + 9 y = 36 \sin 2 t$$ Given that \(y = \frac { 36 } { 5 } \sin 2 t\) is a particular integral of this differential equation,
3. find \(y\) in terms of \(t\),
4. find the speed of \(P\) when \(t = \frac { 1 } { 3 } \pi\).

6. \begin{figure}[h] \end{figure} A railway truck of mass \(M\) approaches the end of a straight horizontal track and strikes a buffer. The buffer is parallel to the track, as shown in Figure 2. The buffer is modelled as a light horizontal spring \(P Q\), which is fixed at the end \(P\). The spring has a natural length \(a\) and modulus of elasticity \(M n ^ { 2 } a\), where \(n\) is a postive constant. At time \(t = 0\), the spring has length \(a\) and the truck strikes the end \(Q\) with speed \(U\). A resistive force whose magnitude is \(M k v\), where \(v\) is the speed of the truck at time \(t\), and \(k\) is a positive constant, also opposes the motion of the truck. At time \(t\), the truck is in contact with the buffer and the compression of the buffer is \(x\).

1. Show that, while the truck is compressing the buffer $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + k \frac { \mathrm {~d} x } { \mathrm {~d} t } + n ^ { 2 } x = 0$$ It is given that \(k = \frac { 5 n } { 2 }\)
2. Find \(x\) in terms of \(U , n\) and \(t\).
3. Find, in terms of \(U\) and \(n\), the greatest value of \(x\).

6. A particle of mass \(m \mathrm {~kg}\) is attached to one end of a light elastic string of natural length a metres and modulus of elasticity 5ma newtons. The other end of the string is attached to a fixed point \(O\) on a smooth horizontal plane. The particle is held at rest on the plane with the string stretched to a length \(2 a\) metres and then released at time \(t = 0\). During the subsequent motion, when the particle is moving with speed \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\), the particle experiences a resistance of magnitude \(4 m v\) newtons. At time \(t\) seconds after the particle is released, the length of the string is ( \(a + x\) ) metres, where \(0 \leqslant x \leqslant a\).

1. Show that, from \(t = 0\) until the string becomes slack, $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 4 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 5 x = 0$$
2. Hence express \(x\) in terms of \(a\) and \(t\).
3. Find the speed of the particle at the instant when the string first becomes slack, giving your answer in the form \(k a\), where \(k\) is a constant to be found correct to 2 significant figures.

4. A particle \(P\) of mass 9 kg moves along the horizontal positive \(x\)-axis under the action of a force directed towards the origin. At time \(t\) seconds, the displacement of \(P\) from \(O\) is \(x\) metres, \(P\) is moving with speed \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and the force has magnitude \(16 x\) newtons. The particle \(P\) is also subject to a resistive force of magnitude \(24 v\) newtons.

1. Show that the equation of motion of \(P\) is $$9 \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 24 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 16 x = 0$$ It is given that the general solution of this differential equation is $$x = \mathrm { e } ^ { - \frac { 4 } { 3 } t } ( A t + B )$$ where \(A\) and \(B\) are arbitrary constants.
   When \(t = \frac { 3 } { 4 } , P\) is travelling towards \(O\) with its maximum speed of \(8 \mathrm { e } ^ { - 1 } \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and \(x = d\).
2. Find the value of \(d\).
3. Find the value of \(x\) when \(t = 0\)

15 In an oscillating system, a particle of mass \(m \mathrm {~kg}\) moves in a horizontal line. Its displacement from its equilibrium position O at time \(t\) seconds is \(x\) metres, its velocity is \(v \mathrm {~ms} ^ { - 1 }\), and it is acted on by a force \(2 m x\) newtons acting towards O as shown in the diagram. \includegraphics[max width=\textwidth, alt={}, center]{b57a2590-84e8-4998-9633-902db861f23a-6_212_914_408_242} Initially, the particle is projected away from O with speed \(1 \mathrm {~ms} ^ { - 1 }\) from a point 2 m from O in the positive direction.

1. 1. Show that the motion is modelled by the differential equation \(\frac { d ^ { 2 } x } { d t ^ { 2 } } + 2 x = 0\).
   2. State the type of motion.
   3. Write down the period of the motion.
   4. Find \(x\) in terms of \(t\).
   5. Find the amplitude of the motion.
2. The motion is now damped by a force \(2 m v\) newtons.
   1. Show that \(\frac { d ^ { 2 } x } { d t ^ { 2 } } + 2 \frac { d x } { d t } + 2 x = 0\).
   2. State, giving a reason, whether the system is under-damped, critically damped or over-damped.
   3. Determine the general solution of this differential equation.
3. Finally, a variable force \(2 m \cos 2 t\) newtons is added, so that the motion is now modelled by the differential equation \(\frac { d ^ { 2 } x } { d t ^ { 2 } } + 2 \frac { d x } { d t } + 2 x = 2 \cos 2 t\).
   1. Find \(x\) in terms of \(t\). In the long term, the particle is seen to perform simple harmonic motion with a period of just over 3 seconds.
   2. Verify that this behaviour is consistent with the answer to part (c)(i).

14 Solve the simultaneous differential equations \(\frac { \mathrm { d } x } { \mathrm {~d} t } + 2 x = 4 y , \quad \frac { \mathrm {~d} y } { \mathrm {~d} t } + 3 x = 5 y\),
given that when \(t = 0 , x = 0\) and \(y = 1\).

13 A particle P of mass \(m\) is fixed to one end of a light spring of natural length \(a\) and modulus of elasticity man \({ } ^ { 2 }\), where \(n > 0\). The other end of the spring is attached to the ceiling of a lift. The lift is at rest and P is hanging vertically in equilibrium.

1. Find, in terms of \(g\) and \(n\), the extension in the spring. At time \(t = 0\) the lift begins to accelerate upwards from rest. At time \(t\), the upward displacement of the lift from its initial position is \(y\) and the extension of the spring is \(x\).
2. Express, in terms of \(g , n , x\) and \(y\), the upward displacement of P from its initial position at time \(t\).
3. Given that \(\ddot { y } = k t\), where \(k\) is a positive constant, express the upward acceleration of \(P\) in terms of \(\ddot { x } , k\) and \(t\).
4. Show that \(x\) satisfies the differential equation $$\ddot { \mathrm { x } } + \mathrm { n } ^ { 2 } \mathrm { x } = \mathrm { kt } + \mathrm { g } .$$
5. Verify that \(\mathrm { x } = \frac { 1 } { \mathrm { n } ^ { 3 } } ( \mathrm { k } n \mathrm { t } + \mathrm { gn } - \mathrm { k } \sin ( \mathrm { nt } ) )\).
6. By considering \(\dot { x }\) comment on the motion of P relative to the ceiling of the lift for all times after the lift begins to move.