1. A company plans to build a new fairground ride. The ride will consist of a capsule that will hold the passengers and the capsule will be attached to a tall tower. The capsule is to be released from rest from a point half way up the tower and then made to oscillate in a vertical line.

The vertical displacement, \(x\) metres, of the top of the capsule below its initial position at time \(t\) seconds is modelled by the differential equation, $$m \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 4 \frac { \mathrm {~d} x } { \mathrm {~d} t } + x = 200 \cos t , \quad t \geqslant 0$$ where \(m\) is the mass of the capsule including its passengers, in thousands of kilograms.
The maximum permissible weight for the capsule, including its passengers, is 30000 N .
Taking the value of \(g\) to be \(10 \mathrm {~ms} ^ { - 2 }\) and assuming the capsule is at its maximum permissible weight,

1. 1. explain why the value of \(m\) is 3
   2. show that a particular solution to the differential equation is $$x = 40 \sin t - 20 \cos t$$
   3. hence find the general solution of the differential equation.
2. Using the model, find, to the nearest metre, the vertical distance of the top of the capsule from its initial position, 9 seconds after it is released.