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 \(( a + y )\).
- Show that \(x + y = \frac { 1 } { 2 } f t ^ { 2 }\).
- 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.
- Find the values of \(A\) and \(B\).
- Find the maximum tension in the spring.
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