A light inextensible string has length 7a. One end of the string is attached to a fixed point \(A\) and the other end of the string is attached to a fixed point \(B\), with \(A\) vertically above \(B\) and \(A B = 5 a\). A particle of mass \(m\) is attached to a point \(P\) on the string where \(A P = 4 a\). The particle moves in a horizontal circle with constant angular speed \(\omega\), with both \(A P\) and \(B P\) taut.
Show that
the tension in \(A P\) is \(\frac { 4 m } { 25 } \left( 9 a \omega ^ { 2 } + 5 g \right)\)
the tension in \(B P\) is \(\frac { 3 m } { 25 } \left( 16 a \omega ^ { 2 } - 5 g \right)\).
The string will break if the tension in it reaches a magnitude of \(4 m g\).
The time for the particle to make one revolution is \(S\).
Show that
$$3 \pi \sqrt { \frac { a } { 5 g } } < S < 8 \pi \sqrt { \frac { a } { 5 g } }$$
State how in your calculations you have used the assumption that the string is light.