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प्रश्न
Repeat the previous exercise if the angle between each pair of springs is 120° initially.
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उत्तर
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,
\[F = kx + \frac{kx}{2} = \frac{3kx}{2} \]
\[ \therefore a = \frac{F}{m} = \frac{3kx}{2m}\]
\[ \Rightarrow \frac{a}{x} = \frac{3k}{2m} = \omega^2 \]
\[ \Rightarrow \omega = \sqrt{\frac{3k}{2m}}\]
\[ \therefore \text { Time period }, T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{2m}{3k}}\]
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