4.10d Second order homogeneous: auxiliary equation method

\includegraphics{figure_2} In a simple model of a shock absorber, a particle \(P\) of mass \(m\) kg is attached to one end of a light elastic horizontal spring. The other end of the spring is fixed at \(A\) and the motion of \(P\) takes place along a fixed horizontal line through \(A\). The spring has natural length \(L\) metres and modulus of elasticity \(2mL\) newtons. The whole system is immersed in a fluid which exerts a resistance on \(P\) of magnitude \(3mv\) newtons, where \(v\) m s\(^{-1}\) is the speed of \(P\) at time \(t\) seconds. The compression of the spring at time \(t\) seconds is \(x\) metres, as shown in Fig. 2.

1. Show that $$\frac{\text{d}^2 x}{\text{d}t^2} + 3\frac{\text{d}x}{\text{d}t} + 2x = 0.$$ [4]

Given that when \(t = 0\), \(x = 2\) and \(\frac{\text{d}x}{\text{d}t} = -4\),

1. find \(x\) in terms of \(t\). [8]
2. Sketch the graph of \(x\) against \(t\). [2]
3. State, with a reason, whether the model is realistic. [1]

A particle \(P\) moves in a straight line. At time \(t\) seconds its displacement from a fixed point \(O\) on the line is \(x\) metres. The motion of \(P\) is modelled by the differential equation $$\frac{\text{d}^2 x}{\text{d}t^2} + 2\frac{\text{d}x}{\text{d}t} + 2x = 12\cos 2t - 6\sin 2t.$$ When \(t = 0\), \(P\) is at rest at \(O\).

1. Find, in terms of \(t\), the displacement of \(P\) from \(O\). [11]
2. Show that \(P\) comes to instantaneous rest when \(t = \frac{\pi}{4}\). [2]
3. Find, in metres to 3 significant figures, the displacement of \(P\) from \(O\) when \(t = \frac{\pi}{4}\). [2]
4. Find the approximate period of the motion for large values of \(t\). [2]

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

1. Show that \(\frac{d^2x}{dt^2} + 2k\frac{dx}{dt} + 4k^2x = 0\). [6]
2. Find an expression, in terms of \(t\), \(k\) and \(L\), for the distance \(AP\) at time \(t\). [8]

A particle of mass \(m\) is attached to one end \(P\) of a light elastic spring \(PQ\), of natural length \(a\) and modulus of elasticity \(man^2\). 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 \(PQ\), with constant acceleration \(f\). At time \(t\), the displacement of the particle in the direction \(PQ\) from its initial position is \(x\) and the length of the spring is \((a + y)\).

1. Show that \(x + y = \frac{1}{2}ft^2\). [2]
2. Hence show that $$\frac{d^2x}{dt^2} + n^2x = \frac{1}{2}n^2ft^2.$$ [6]

You are given that the general solution of this differential equation is $$x = A\cos nt + B\sin nt + \frac{1}{2}ft^2 - \frac{f}{n^2},$$ where \(A\) and \(B\) are constants.

1. Find the values of \(A\) and \(B\). [6]
2. Find the maximum tension in the spring. [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 \(2m\omega^2a\), 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 \(2m\omega v\), 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 \(\frac{\mathrm{d}^2x}{\mathrm{d}t^2} + 2\omega \frac{\mathrm{d}x}{\mathrm{d}t} + 2\omega^2x = 0\). [5] Given that the solution of this differential equation is \(x = e^{-\omega t}(A \cos \omega t + B \sin \omega t)\), where \(A\) and \(B\) are constants,
2. find \(A\) and \(B\). [4]
3. Find an expression for the time at which \(P\) first comes to rest. [3]

\includegraphics{figure_2} 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 \(PQ\), which is fixed at the end \(P\). The spring has a natural length \(a\) and modulus of elasticity \(Mn^2a\), where \(n\) is a positive 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 \(Mkv\), 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 $$\frac{\text{d}^2x}{\text{d}t^2} + k\frac{\text{d}x}{\text{d}t} + n^2x = 0$$ (4)

It is given that \(k = \frac{5n}{2}\)

1. Find \(x\) in terms of \(U\), \(n\) and \(t\). (7)

1. Find, in terms of \(U\) and \(n\), the greatest value of \(x\). (5)

A particle of mass \(m\) 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 \(2a\) metres and then released at time \(t = 0\). During the subsequent motion, when the particle is moving with speed \(v\) m s\(^{-1}\), the particle experiences a resistance of magnitude \(4mv\) 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, $$\frac{\mathrm{d}^2 x}{\mathrm{d}t^2} + 4\frac{\mathrm{d}x}{\mathrm{d}t} + 5x = 0$$ [3]
2. Hence express \(x\) in terms of \(a\) and \(t\). [6]
3. Find the speed of the particle at the instant when the string first becomes slack, giving your answer in the form \(ka\), where \(k\) is a constant to be found correct to 2 significant figures. [4]

An elastic string spring of modulus \(2mg\) and natural length \(l\) is fixed at one end. To the other end is attached a mass \(m\) which is allowed to hang in equilibrium. The mass is then pulled vertically downwards through a distance \(l\) and released from rest. The air resistance is modelled as having magnitude \(2m\omega v\), where \(v\) is the speed of the particle and \(\omega = \sqrt{\frac{g}{l}}\). The particle is at distance \(x\) from its equilibrium position at time \(t\).

1. Show that \(\frac{\mathrm{d}^2 x}{\mathrm{d} t^2} + 2\omega \frac{\mathrm{d} x}{\mathrm{d} t} + 2\omega^2 x = 0\). [7]
2. Find the general solution of this differential equation. [4]
3. Hence find the period of the damped harmonic motion. [1]

A particle moves in a plane in such a way that its position vector \(\mathbf{r}\) metres at time \(t\) seconds satisfies the differential equation $$\frac{d^2\mathbf{r}}{dt^2} - 2\frac{d\mathbf{r}}{dt} = 0$$ When \(t = 0\), the particle is at the origin and is moving with velocity \((4i + 2j)\) m s\(^{-1}\). Find \(\mathbf{r}\) in terms of \(t\). [7]

A particle \(P\) moves in the \(x\)-\(y\) plane and has position vector \(\mathbf{r}\) metres at time \(t\) seconds. It is given that \(\mathbf{r}\) satisfies the differential equation $$\frac{\mathrm{d}^2\mathbf{r}}{\mathrm{d}t^2} = 2\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}.$$ When \(t = 0\), \(P\) is at the point with position vector \(3\mathbf{i}\) metres and is moving with velocity \(\mathbf{j}\) m s\(^{-1}\).

1. Find \(\mathbf{r}\) in terms of \(t\). [8]
2. Describe the path of \(P\), giving its cartesian equation. [2]

Find the general solution of the differential equation $$\frac{d^2y}{dx^2} - c\frac{dy}{dx} + 8y = e^{3x}.$$ [6]

Find the general solution of the differential equation $$\frac{d^2y}{dx^2} - 8\frac{dy}{dx} + 16y = 4x.$$ [7]

Find the general solution of the differential equation $$\frac{d^2y}{dx^2} + 4\frac{dy}{dx} + 5y = 65 \sin 2x.$$ [9]

The variables \(x\) and \(y\) satisfy the differential equation $$\frac{d^2y}{dx^2} - 6\frac{dy}{dx} + 9y = e^{3x}.$$

1. Find the complementary function. [3]
2. Explain briefly why there is no particular integral of either of the forms \(y = ke^{3x}\) or \(y = kxe^{3x}\). [1]
3. Given that there is a particular integral of the form \(y = kx^2e^{3x}\), find the value of \(k\). [5]

The variables \(x\) and \(y\) satisfy the differential equation $$\frac{\text{d}^2y}{\text{d}x^2} + 16y = 8\cos 4x.$$

1. Find the complementary function of the differential equation. [2]
2. Given that there is a particular integral of the form \(y = px\sin 4x\), where \(p\) is a constant, find the general solution of the equation. [6]
3. Find the solution of the equation for which \(y = 2\) and \(\frac{\text{d}y}{\text{d}x} = 0\) when \(x = 0\). [4]

Find the general solution of the differential equation $$\frac{\mathrm{d}^2y}{\mathrm{d}x^2} - 3\frac{\mathrm{d}y}{\mathrm{d}x} - 4y = \cos 2x + 5x$$ [9 marks]

A tourist decides to do a bungee jump from a bridge over a river. One end of an elastic rope is attached to the bridge and the other end of the elastic rope is attached to the tourist. The tourist jumps off the bridge. At time \(t\) seconds after the tourist reaches their lowest point, their vertical displacement is \(x\) metres above a fixed point 30 metres vertically above the river. When \(t = 0\)

• \(x = -20\)
• the velocity of the tourist is \(0\text{ms}^{-1}\)
• the acceleration of the tourist is \(13.6\text{ms}^{-2}\)

In the subsequent motion, the elastic rope is assumed to remain taut so that the vertical displacement of the tourist can be modelled by the differential equation $$5k\frac{d^2x}{dt^2} + 2k\frac{dx}{dt} + 17x = 0 \quad t \geq 0$$ where \(k\) is a positive constant.

1. Determine the value of \(k\) [2]
2. Determine the particular solution to the differential equation. [7]
3. Hence find, according to the model, the vertical height of the tourist above the river 15 seconds after they have reached their lowest point. [2]
4. Give a limitation of the model. [1]

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? [1]
   2. Explain briefly why modelling the motion of the door as simple harmonic is unlikely to be realistic. [1]
2. Find the range of values of \(\lambda\) for which the model predicts that the door will never pass through the equilibrium position. [2]
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. [1]

In this question take \(g = 10\). A particle P of mass 0.15 kg is attached to one end of a light elastic string of modulus of elasticity 13.5 N and natural length 0.45 m. The other end of the string is attached to a fixed point O. The particle P rests in equilibrium at a point A with the string vertical.

1. Show that the distance OA is 0.5 m. [3]

At time \(t = 0\), P is projected vertically downwards from A with a speed of 1.25 m s\(^{-1}\). Throughout the subsequent motion, P experiences a variable resistance \(R\) newtons which is of magnitude 0.6 times its speed (in m s\(^{-1}\)).

1. Given that the downward displacement of P from A at time \(t\) seconds is \(x\) metres, show that, while the string remains taut, \(x\) satisfies the differential equation $$\frac{d^2x}{dt^2} + 4\frac{dx}{dt} + 200x = 0.$$ [3]
2. Verify that \(x = \frac{5}{56}e^{-2t}\sin(14t)\). [6]
3. Determine whether the string becomes slack during the motion. [5]

The sequence \(u_0, u_1, u_2, \ldots\) satisfies the recurrence relation \(u_{n+2} - 3u_{n+1} - 10u_n = 24n - 10\).

1. Determine the general solution of the recurrence relation. [6]
2. Hence determine the particular solution of the recurrence relation for which \(u_0 = 6\) and \(u_1 = 10\). [3]
3. Show, by direct calculation, that your solution in part (b) gives the correct value for \(u_2\). [1]

The sequence \(v_0, v_1, v_2, \ldots\) is defined by \(v_n = \frac{u_n}{p^n}\) for some constant \(p\), where \(u_n\) denotes the particular solution found in part (b). You are given that \(v_n\) converges to a finite non-zero limit, \(q\), as \(n \to \infty\).

1. Determine \(p\) and \(q\). [4]

1. Find the general solution of $$u_n = 8u_{n-1} - 16u_{n-2}, \quad n \geq 2. \quad (*)$$ [4]

A new sequence \(v_n\) is defined by \(v_n = \frac{u_n}{u_{n-1}}\) for \(n \geq 1\).

1. (A) Use (*) to show that \(v_n = 8 - \frac{16}{v_{n-1}}\) for \(n \geq 2\). [2] (B) Deduce that if \(v_n\) tends to a limit then it must be 4. [2]
2. Use your general solution in part (i) to show that \(\lim_{n \to \infty} v_n = 4\). [3]
3. Deduce the value of \(\lim_{n \to \infty} \left(\frac{u_n}{u_{n-2}}\right)\). [1]