Prove SHM and find period: given force or equation of motion directly

Two particles \(A\) and \(B\), of equal mass \(m\), are connected by a light elastic string of natural length \(a\) and modulus of elasticity \(4 m g\). Particle \(A\) rests on a rough horizontal table at a distance \(a\) from the edge of the table. The string passes over a small smooth pulley \(P\) fixed at the edge of the table. At time \(t = 0 , B\) is released from rest at \(P\) and falls vertically. At time \(t , B\) has fallen a distance \(x\), without \(A\) slipping (see diagram). Show that $$\ddot { x } = - \frac { g } { a } ( 4 x - a ) .$$ Deduce that, while \(A\) does not slip, \(B\) moves in simple harmonic motion and identify the centre of the motion. Given that the coefficient of friction between \(A\) and the table is \(\frac { 1 } { 3 }\), find the value of \(x\) when \(A\) starts to slip, and the corresponding value of \(t\), expressing this answer in the form \(k \sqrt { } \left( \frac { a } { g } \right)\). Give the value of \(k\) correct to 3 decimal places.

3 \includegraphics[max width=\textwidth, alt={}, center]{f6887893-66c5-40df-ba8d-9439a5c268eb-3_351_314_255_918} A spring balance is modelled by a vertical light elastic spring \(A B\), of natural length 0.25 m and modulus of elasticity \(\lambda \mathrm { N }\). The bottom end \(B\) of the spring is fixed, and the top end \(A\) is attached to a small tray of mass 0.1 kg which is free to move vertically (see diagram). When in the equilibrium position, \(A B = 0.24 \mathrm {~m}\). Show that \(\lambda = 25\). The tray is pushed down by 0.02 m to the point \(C\) and released from rest. At time \(t\) seconds after release the displacement of the tray from the equilibrium position is \(x \mathrm {~m}\). Show that $$\ddot { x } = - 1000 x .$$ Find the time taken for the tray to move a distance of 0.03 m from \(C\).

6 A particle \(P\) of mass 0.1 kg moves in a straight line on a smooth horizontal surface. A force of \(( 0.36 - 0.144 x ) \mathrm { N }\) acts on \(P\) in the direction from \(O\) to \(P\), where \(x \mathrm {~m}\) is the displacement of \(P\) from a point \(O\) on the surface at time \(t \mathrm {~s}\).

1. By using the substitution \(x = y + 2.5\), or otherwise, show that \(P\) moves with simple harmonic motion of period 5.24 s , correct to 3 significant figures. The maximum value of \(x\) during the motion is 3 .
2. Write down the amplitude of \(P\) 's motion and find the two possible values of \(x\) for which \(P\) 's speed is \(0.48 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
3. On each of the first two occasions when \(P\) has speed \(0.48 \mathrm {~m} \mathrm {~s} ^ { - 1 } , P\) is moving towards \(O\). Find the time interval between
   1. these first two occasions,
   2. the second and third occasions when \(P\) has speed \(0.48 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).

3 A bungee jumper of mass 75 kg is connected to a fixed point A by a light elastic rope with stiffness \(300 \mathrm { Nm } ^ { - 1 }\). The jumper starts at rest at A and falls vertically. The lowest point reached by the jumper is 40 m vertically below A. Air resistance may be neglected.

1. Find the natural length of the rope.
2. Show that, when the rope is stretched and its extension is \(x\) metres, \(\ddot { x } + 4 x = 9.8\). The substitution \(y = x - c\), where \(c\) is a constant, transforms this equation to \(\ddot { y } = - 4 y\).
3. Find \(c\), and state the maximum value of \(y\).
4. Using standard simple harmonic motion formulae, or otherwise, find
   (A) the maximum speed of the jumper,
   (B) the maximum deceleration of the jumper.
5. Find the time taken for the jumper to fall from A to the lowest point.

2 A light elastic string has natural length \(a\) and modulus of elasticity \(k m g\), where \(k > 2\). One end of the string is attached to a fixed point O . A particle P of mass \(m\) is attached to the other end of the string. P is held at rest a distance \(\frac { 3 } { 2 } a\) vertically below O . At time \(t\) after P is released, its vertical distance below O is \(y\).

1. Show that, while the string is in tension, the equation of motion of P is given by the differential equation \(\frac { d ^ { 2 } y } { d t ^ { 2 } } = ( k + 1 ) g - \frac { k g } { a } y\). A student transforms the differential equation in part (a) into the standard SHM equation \(\frac { d ^ { 2 } x } { d t ^ { 2 } } = - \omega ^ { 2 } x\).
2. - Find an expression for \(x\) in terms of \(y , k\) and \(a\).

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]

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]

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]

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]