One end of a light elastic string, of natural length \(0 \cdot 2 \mathrm {~m}\) and modulus of elasticity \(5 g \mathrm {~N}\), is attached to a fixed point \(O\). The other end is attached to a particle of mass 4 kg . The particle hangs in equilibrium vertically below \(O\).
Show that the extension of the string is 0.16 m .
The particle is pulled down vertically and held at rest so that the extension of the string is 0.28 m . The particle is then released. Determine the speed of the particle as it passes through the equilibrium position.
At time \(t = 0\) seconds, a particle \(A\) has position vector \(( 6 \mathbf { i } + 21 \mathbf { j } - 8 \mathbf { k } )\) m relative to a fixed origin O and is moving with constant velocity \(( 3 \mathbf { i } - \mathbf { j } + 4 \mathbf { k } ) \mathrm { ms } ^ { - 1 }\).
Write down the position vector of particle \(A\) at time \(t\) seconds and hence find the distance \(O A\) when \(t = 5\).
The position vector, \(\mathbf { r } _ { B }\) metres, of another particle \(B\) at time \(t\) seconds is given by
$$\mathbf { r } _ { B } = 3 \sin \left( \frac { t } { 2 } \right) \mathbf { i } - 3 \cos \left( \frac { t } { 2 } \right) \mathbf { j } + 5 \mathbf { k } .$$
(i) Show that \(B\) is moving with constant speed.
(ii) Determine the smallest value of \(t\) such that particles \(A\) and \(B\) are moving perpendicular to each other.