12 A particle P of mass \(m\) is fixed to one end of a light elastic string of natural length \(l\) and modulus of elasticity 12 mg . The other end of the string is attached to a fixed point O . Particle P is held next to O and then released from rest.
- Show that P next comes instantaneously to rest when the length of the string is \(\frac { 3 } { 2 } l\).
The string first becomes taut at time \(t = 0\). At time \(t \geqslant 0\), the length of the string is \(l + x\), where \(x\) is the extension in the string.
- Show that when the string is taut, \(x\) satisfies the differential equation
$$\ddot { \mathrm { x } } + \omega ^ { 2 } \mathrm { x } = \mathrm { g } \text {, where } \omega ^ { 2 } = \frac { 12 \mathrm {~g} } { \mathrm { I } } \text {. }$$
- By using the substitution \(x = y + \frac { g } { \omega ^ { 2 } }\), solve the differential equation to show that the time when the string first becomes slack satisfies the equation
$$\cos \omega \mathrm { t } - \sqrt { \mathrm { k } } \sin \omega \mathrm { t } = 1$$
where \(k\) is an integer to be determined.