Question

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

Answer 1

Explanation:

• We know that time period of oscillation, $T=\frac{2 \pi}{\omega}$, where ω is the angular velocity.  
• Let us consider the particle ‘m’ is pushed against ‘C’ by distance ‘x’ when the angle between each pair of springs is 120° initially.  Then, the force due to springs A and B is  $=\sqrt{2\left(\frac{\mathrm{kx}}{2}\right)^{2}+2\left(\frac{\mathrm{kx}}{2}\right)\left(\frac{k x}{2}\right) \cos 120^{\circ}}.$
• Thus, the net force due to A and B is $k \times \frac{x}{2}$ and the total force acting on the particle ‘m’ is  $F=3 k * \frac{x}{2}, \text { As } a=\frac{F}{m},\left(\frac{3 k}{2 m}\right)=\omega^{2}$ which implies that $\omega=\sqrt{\frac{3 \mathrm{k}}{2 \mathrm{m}}}.$Therefore, on substituting the known values in the time period formula, we get time period    $\mathrm{T}=2 \pi \sqrt{\frac{2 \mathrm{m}}{3 \mathrm{k}}}.$