4.10f Simple harmonic motion: x'' = -omega^2 x

\includegraphics{figure_6} A compound pendulum consists of a uniform rod \(AB\) of length 1 m and mass 3 kg, a particle of mass 1 kg attached to the rod at \(A\) and a circular disc of radius \(\frac{1}{5}\) m, mass 6 kg and centre \(C\). The end \(B\) of the rod is rigidly attached to a point on the circumference of the disc in such a way that \(ABC\) is a straight line. The pendulum is initially at rest with \(B\) vertically below \(A\) and it is free to rotate in a vertical plane about a smooth fixed horizontal axis passing through the point \(P\) on the rod where \(AP = x\) m and \(x < \frac{1}{3}\) (see diagram).

1. Show that the moment of inertia of the pendulum about the axis of rotation is \((10x^2 - 19x + 12)\) kg m\(^2\). [6]

The pendulum is making small oscillations about the equilibrium position, such that at time \(t\) seconds the angular displacement that the pendulum makes with the downward vertical is \(\theta\) radians.

1. Find the angular acceleration of the pendulum, in terms of \(x\), \(g\) and \(\theta\). [4]
2. Show that the motion is approximately simple harmonic, and show that the approximate period of oscillations, in seconds, is given by \(2\pi\sqrt{\frac{20x^2 - 38x + 24}{(19-20x)g}}\). [2]
3. Hence find the value of \(x\) for which the approximate period of oscillations is least. [3]

A uniform lamina \(ABC\) of mass \(m\) is in the shape of an isosceles triangle with \(AB = AC = 5a\) and \(BC = 8a\).

1. Show, using integration, that the moment of inertia of the lamina about an axis through \(A\), parallel to \(BC\), is \(\frac{9}{2}ma^2\). [6]

The foot of the perpendicular from \(A\) to \(BC\) is \(D\). The lamina is free to rotate in a vertical plane about a fixed smooth horizontal axis which passes through \(D\) and is perpendicular to the plane of the lamina. The lamina is released from rest when \(DA\) makes an angle \(\alpha\) with the downward vertical. It is given that the moment of inertia of the lamina about an axis through \(D\), perpendicular to \(BC\) and in the plane of the lamina, is \(\frac{8}{3}ma^2\).

1. Find the angular acceleration of the lamina when \(DA\) makes an angle \(\theta\) with the downward vertical. [8]

Given that \(\alpha\) is small,

1. find an approximate value for the period of oscillation of the lamina about the vertical. [2]

A uniform square lamina \(ABCD\) of side \(a\) and mass \(m\) is free to rotate in vertical plane about a horizontal axis through \(A\). The axis is perpendicular to the plane of the lamina. The lamina is released from rest when \(t = 0\) and \(AC\) makes a small angle with the downward vertical through \(A\).

1. Show that the moment of inertia of the lamina about the axis is \(\frac{2}{3}ma^2\). [3]
2. Show that the motion of the lamina is approximately simple harmonic. [5]
3. Find the time \(t\) when \(AC\) is first vertical. [2]

A particle \(P\) moves in the \(x\)-\(y\) plane and has position vector \(\mathbf{r}\) metres relative to a fixed origin \(O\) at time \(t\) s. Given that \(\mathbf{r}\) satisfies the vector differential equation $$\frac{d^2\mathbf{r}}{dt^2} + 9\mathbf{r} = 8\sin t \mathbf{i}$$ and that when \(t = 0\) s, \(P\) is at \(O\) and moving with velocity \((\mathbf{i} + 3\mathbf{j})\) m s\(^{-1}\),

1. find \(\mathbf{r}\) at time \(t\). [11]
2. Hence find when \(P\) next returns to \(O\). [2]

In this question use \(g = 10 \text{ m s}^{-2}\) A light spring is attached to the base of a long tube and has a mass \(m\) attached to the other end, as shown in the diagram. The tube is filled with oil. When the compression of the spring is \(c\) metres, the thrust in the spring is \(9mc\) newtons. \includegraphics{figure_14} The mass is held at rest in a position where the compression of the spring is \(\frac{20}{9}\) metres. The mass is then released from rest. During the subsequent motion the oil causes a resistive force of \(6mv\) newtons to act on the mass, where \(v \text{ m s}^{-1}\) is the speed of the mass. At time \(t\) seconds after the mass is released, the displacement of the mass above its starting position is \(x\) metres.

1. Find \(x\) in terms of \(t\). [10 marks]
2. State, giving a reason, the type of damping which occurs. [1 mark]

The displacement of a particle from its equilibrium position is \(x\) metres at time \(t\) seconds. The motion of the particle obeys the differential equation $$\frac{d^2x}{dt^2} = -9x$$ Calculate the period of its motion in seconds. Circle your answer. [1 mark] \(\frac{\pi}{9}\) \(\quad\) \(\frac{2\pi}{9}\) \(\quad\) \(\frac{\pi}{3}\) \(\quad\) \(\frac{2\pi}{3}\)

In this question use \(g\) as \(10\,\text{m}\,\text{s}^{-2}\) A smooth plane is inclined at \(30°\) to the horizontal. The fixed points \(A\) and \(B\) are 3.6 metres apart on the line of greatest slope of the plane, with \(A\) higher than \(B\) A particle \(P\) of mass 0.32 kg is attached to one end of each of two light elastic strings. The other ends of these strings are attached to the points \(A\) and \(B\) respectively. The particle \(P\) moves on a straight line that passes through \(A\) and \(B\) \includegraphics{figure_2} The natural length of the string \(AP\) is 1.4 metres. When the extension of the string \(AP\) is \(e_A\) metres, the tension in the string \(AP\) is \(7e_A\) newtons. The natural length of the string \(BP\) is 1 metre. When the extension of the string \(BP\) is \(e_B\) metres, the tension in the string \(BP\) is \(9e_B\) newtons. The particle \(P\) is held at the point between \(A\) and \(B\) which is 0.2 metres from its equilibrium position and lower than its equilibrium position. The particle \(P\) is then released from rest. At time \(t\) seconds after \(P\) is released, its displacement towards \(B\) from its equilibrium position is \(x\) metres.

1. Show that during the subsequent motion the object satisfies the equation $$\ddot{x} + 50x = 0$$ Fully justify your answer. [5 marks]
2. The experiment is repeated in a large tank of oil. During the motion the oil causes a resistive force of \(kv\) newtons to act on the particle, where \(v\,\text{m}\,\text{s}^{-1}\) is the speed of the particle. The oil causes critical damping to occur.
   1. Show that \(k = \frac{16\sqrt{2}}{5}\) [3 marks]
   2. Find \(x\) in terms of \(t\), giving your answer in exact form. [6 marks]
   3. Calculate the maximum speed of the particle. [5 marks]

A particle, \(P\), of mass \(M\) is released from rest and moves along a horizontal straight line through a point \(O\). When \(P\) is at a displacement of \(x\) metres from \(O\), moving with a speed \(v\) ms\(^{-1}\), a force of magnitude \(|8Mx|\) acts on the particle directed towards \(O\). A resistive force, of magnitude \(4Mv\), also acts on \(P\).

1. Initially \(P\) is held at rest at a displacement of 1 metre from \(O\). Describe completely the motion of \(P\) after it is released. Fully justify your answer. [8 marks]
2. It is decided to alter the resistive force so that the motion of \(P\) is critically damped. Determine the magnitude of the resistive force that will produce critically damped motion. [4 marks]

A bungee jumper of mass \(m\) kg is attached to an elastic rope. The other end of the rope is attached to a fixed point. The bungee jumper falls vertically from the fixed point. At time \(t\) seconds after the rope first becomes taut, the extension of the rope is \(x\) metres and the speed of the bungee jumper is \(v\) m s\(^{-1}\)

1. A model for the motion while the rope remains taut assumes that the forces acting on the bungee jumper are • the weight of the bungee jumper • a tension in the rope of magnitude \(kx\) newtons • an air resistance force of magnitude \(Rv\) newtons where \(k\) and \(R\) are constants such that \(4km > R^2\)
   1. Show that this model gives the result $$x = e^{-\frac{Rt}{2m}} \left( A \cos \frac{\sqrt{4km - R^2}}{2m} t + B \sin \frac{\sqrt{4km - R^2}}{2m} t \right) + \frac{mg}{k}$$ where \(A\) and \(B\) are constants, and \(g\) m s\(^{-2}\) is the acceleration due to gravity. You do not need to find the value of \(A\) or the value of \(B\) [6 marks]
   2. It is also given that: \(k = 16\) \(R = 20\) \(m = 62.5\) \(g = 9.8\) m s\(^{-2}\) and that the speed of the bungee jumper when the rope becomes taut is 14 m s\(^{-1}\) Show that, to the nearest integer, \(A = -38\) and \(B = 16\) [6 marks]
2. A second, simpler model assumes that the air resistance is zero. The values of \(k\), \(m\) and \(g\) remain the same. Find an expression for \(x\) in terms of \(t\) according to this simpler model, giving the values of all constants to two significant figures. [4 marks]

The movement of a particle is described by the simple harmonic equation $$\ddot{x} = -25x$$ where \(x\) metres is the displacement of the particle at time \(t\) seconds, and \(\ddot{x}\) m s\(^{-2}\) is the acceleration of the particle. The maximum displacement of the particle is 9 metres. Find the maximum speed of the particle. Circle your answer. [1 mark] \(15\) m s\(^{-1}\) \quad\quad \(45\) m s\(^{-1}\) \quad\quad \(75\) m s\(^{-1}\) \quad\quad \(135\) m s\(^{-1}\)

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]

A small object is attached to a spring and performs oscillations in a vertical line. The displacement of the object at time \(t\) seconds is denoted by \(x\) cm. Preliminary observations suggest that the object performs simple harmonic motion (SHM) with a period of 2 seconds about the point at which \(x = 0\).

1. 1. Write down a differential equation to model this motion. [3]
   2. Give the general solution of the differential equation in part (i) (A). [1]

Subsequent observations indicate that the object's motion would be better modelled by the differential equation $$\frac{d^2x}{dt^2} + 2k \frac{dx}{dt} + (k^2 + 9)x = 0 \qquad (*)$$ where \(k\) is a positive constant.

1. 1. Obtain the general solution of (*). [3]
   2. State two ways in which the motion given by this model differs from that in part (i). [2]

The amplitude of the object's motion is observed to reduce with a scale factor of 0.98 from one oscillation to the next.

1. Find the value of \(k\). [3]

At the start of the object's motion, \(x = 0\) and the velocity is 12 cm s\(^{-1}\) in the positive \(x\) direction.

1. Find an equation for \(x\) as a function of \(t\). [4]
2. Without doing any further calculations, explain why, according to this model, the greatest distance of the object from its starting point in the subsequent motion will be slightly less than 4 cm. [2]

\includegraphics{figure_9} A particle P of mass \(m\) is joined to a fixed point O by a light inextensible string of length \(l\). P is released from rest with the string taut and making an acute angle \(\alpha\) with the downward vertical, as shown in Fig. 9. At a time \(t\) after P is released the string makes an angle \(\theta\) with the downward vertical and the tension in the string is \(T\). Angles \(\alpha\) and \(\theta\) are measured in radians.

1. Show that $$\left(\frac{\mathrm{d}\theta}{\mathrm{d}t}\right)^2 = \frac{2g}{l}\cos\theta + k_1,$$ where \(k_1\) is a constant to be determined in terms of \(g\), \(l\) and \(\alpha\). [4]
2. Show that $$T = 3mg\cos\theta + k_2,$$ where \(k_2\) is a constant to be determined in terms of \(m\), \(g\) and \(\alpha\). [3]

It is given that \(\alpha\) is small enough for \(\alpha^2\) to be negligible.

1. Find, in terms of \(m\) and \(g\), the approximate tension in the string. [2]
2. Show that the motion of P is approximately simple harmonic. [3]

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. [3]

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 \(x\).

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

A particle P of mass \(3m\) kg is attached to one end of a light elastic string of modulus of elasticity \(4mg\) N and natural length 0.4 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. Find the distance OA. [2]

At time \(t = 0\) seconds, P is given a speed of \(2.5 \text{ m s}^{-1}\) vertically downwards from A.

1. Show that P initially performs simple harmonic motion with amplitude \(a\) m, where \(a\) is to be determined correct to 3 significant figures. [5]
2. Determine the smallest distance between P and O in the subsequent motion. [3]

\includegraphics{figure_7} A particle P of mass \(m\) is attached to one end of a light elastic string of natural length \(6a\) and modulus of elasticity \(3mg\). The other end of the string is fixed to a point O on a smooth plane, which is inclined at an angle of \(30°\) to the horizontal. The string lies along a line of greatest slope of the plane and P rests in equilibrium on the inclined plane at a point A, as shown in Fig. 7. P is now pulled a further distance \(2a\) down the line of greatest slope through A and released from rest. At time \(t\) later, the displacement of P from A is \(x\), where the positive direction of \(x\) is down the plane.

1. Show that, until the string slackens, \(x\) satisfies the differential equation $$\frac{d^2x}{dt^2} + \frac{gx}{2a} = 0.$$ [6]
2. Determine, in terms of \(a\) and \(g\), the time at which the string slackens. [5]
3. Find, in terms of \(a\) and \(g\), the speed of P when the string slackens. [2]

\(A\) is a fixed point on a smooth horizontal surface. A particle \(P\) is initially held at \(A\) and released from rest. It subsequently performs simple harmonic motion in a straight line on the surface. After its release it is next at rest after \(0.2\) seconds at point \(B\) whose displacement is \(0.2\) m from \(A\). The point \(M\) is halfway between \(A\) and \(B\). The displacement of \(P\) from \(M\) at time \(t\) seconds after release is denoted by \(x\) m.

1. Sketch a graph of \(x\) against \(t\) for \(0 \leq t \leq 0.4\). [4]
2. Find the displacement of \(P\) from \(M\) at \(0.75\) seconds after release. [2]

A swing door is a door to a room which is closed when in equilibrium but which can be pushed open from either side and which can swing both ways, into or out of the room, and through the equilibrium position. The door is sprung so that when displaced from the equilibrium position it will swing back towards it. The extent to which the door is open at any time, \(t\) seconds, is measured by the angle at the hinge, \(\theta\), which the plane of the door makes with the plane of the equilibrium position. See the diagram below. In an initial model of the motion of a certain swing door it is suggested that \(\theta\) satisfies the following differential equation. $$4\frac{\mathrm{d}^2\theta}{\mathrm{d}t^2} + 25\theta = 0 \quad (*)$$

1. 1. Write down the general solution to (*). [2]
   2. With reference to the behaviour of your solution in part (a)(i) explain briefly why the model using (*) is unlikely to be realistic. [1]

In an improved model of the motion of the door an extra term is introduced to the differential equation so that it becomes $$4\frac{\mathrm{d}^2\theta}{\mathrm{d}t^2} + \lambda\frac{\mathrm{d}\theta}{\mathrm{d}t} + 25\theta = 0 \quad (\dagger)$$ where \(\lambda\) is a positive constant.

1. In the case where \(\lambda = 16\) the door is held open at an angle of \(0.9\) radians and then released from rest at time \(t = 0\).
   1. Find, in a real form, the general solution of (\(\dagger\)). [3]
   2. Find the particular solution of (\(\dagger\)). [4]
   3. With reference to the behaviour of your solution found in part (b)(ii) explain briefly how the extra term in (\(\dagger\)) improves the model. [2]
2. Find the value of \(\lambda\) for which the door is critically damped. [2]

One end of a light spring of length 0.5 m is attached to a fixed point \(F\). A particle \(P\) of mass 2.5 kg is attached to the other end of the spring and hangs in equilibrium 0.55 m below \(F\). Another particle \(Q\), of mass 1.5 kg, is attached to \(P\), without moving it, and both particles are then released.

1. Show that the modulus of elasticity of the spring is 250 N. [2]
2. Prove that the motion is simple harmonic. [4]
3. Find
   1. the amplitude of the motion, [1]
   2. the greatest speed of the particles, [1]
   3. the period of the motion, [1]
   4. the time taken for the spring to be extended by 0.1 m for the first time. [3]

A light elastic string of natural length \(2a\) and modulus of elasticity \(\lambda\) is stretched between two points \(A\) and \(B\), which are \(3a\) apart on a smooth horizontal table. A particle of mass \(m\) is attached to the mid-point of the string, pulled aside to \(A\) and released.

1. Prove that, while one part of the string is taut and the other part is slack, the particle is describing simple harmonic motion. [2]
2. Find the speed of the particle when the slack part of the string becomes taut. [2]
3. Prove that the total time for the particle to reach the mid-point of the string for the first time is $$\sqrt{\frac{ma}{\lambda}} \left( \frac{\pi}{3} + \frac{1}{\sqrt{2}} \sin^{-1} \frac{1}{\sqrt{7}} \right).$$ [8]

11 Answer only one of the following two alternatives.
EITHER
A particle \(P\) of mass \(m\) is suspended from a fixed point by a light elastic string of natural length \(l\), and hangs in equilibrium. The particle is pulled vertically down to a position where the length of the string is \(\frac { 13 } { 7 } l\). The particle is released from rest in this position and reaches its greatest height when the length of the string is \(\frac { 11 } { 7 } l\).

1. Show that the modulus of elasticity of the string is \(\frac { 7 } { 5 } \mathrm { mg }\).
2. Show that \(P\) moves in simple harmonic motion about the equilibrium position and state the period of the motion.
3. Find the time after release when the speed of \(P\) is first equal to half of its maximum value.
   OR
   For a random sample of 12 observations of pairs of values \(( x , y )\), the equation of the regression line of \(y\) on \(x\) and the equation of the regression line of \(x\) on \(y\) are $$y = b x + 4.5 \quad \text { and } \quad x = a y + c$$ respectively, where \(a , b\) and \(c\) are constants. The product moment correlation coefficient for the sample is 0.6 .
4. Test, at the \(5 \%\) significance level, whether there is evidence of positive correlation between the variables.
5. Given that \(b - a = 0.5\), find the values of \(a\) and \(b\).
6. Given that the sum of the \(x\)-values in the sample data is 66, find the value of \(c\) and sketch the two regression lines on the same diagram. For each of the 12 pairs of values of \(( x , y )\) in the sample, another variable \(z\) is considered, where \(z = 5 y\).
7. State the coefficient of \(x\) in the equation of the regression line of \(z\) on \(x\) and find the value of the product moment correlation coefficient between \(x\) and \(z\), justifying your answer.