Question

Repeat the previous exercise if the angle between each pair of springs is 120° initially.

Answer 1

As the particle is pushed against the spring C by the distance x, it experiences a force of magnitude kx.

If the angle between each pair of the springs is 120˚ then the net force applied by the springs A and B is given as,

\[\sqrt{\left( \frac{kx}{2} \right)^2 + \left( \frac{kx}{2} \right)^2 + 2\left( \frac{kx}{2} \right)\left( \frac{kx}{2} \right)  \cos  120^\circ}   = \frac{kx}{2}\]

Total resultant force \[\left( F \right)\] acting on mass m will be,

\[F = kx + \frac{kx}{2} = \frac{3kx}{2}                          \] 

\[ \therefore a = \frac{F}{m} = \frac{3kx}{2m}\] 

\[ \Rightarrow \frac{a}{x} = \frac{3k}{2m} =  \omega^2 \] 

\[ \Rightarrow \omega = \sqrt{\frac{3k}{2m}}\] 

\[ \therefore \text { Time  period },   T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{2m}{3k}}\]