Question
Repeat the previous exercise if the angle between each pair of springs is 120° initially.
✓
Answer
In this case, if the particle ‘m’ is pushed against ’C’ a by distance ‘x’. Total resultant force acting on man ‘m’ is given by,$\text{F}=\text{kx}+\frac{\text{kx}}{2}=\frac{3\text{kx}}{2}$
Because net force $\text{A }\&\text{ B}=\sqrt{\Big(\frac{\text{kx}}{2}\Big)^2+\Big(\frac{\text{kx}}{2}\Big)^2+2\Big(\frac{\text{kx}}{2}\Big)\Big(\frac{\text{kx}}{2}\Big)\cos120^\circ}=\frac{\text{kx}}{2}$$\therefore\text{a}=\frac{\text{F}}{\text{m}}=\frac{3\text{kx}}{2\text{m}}$
$\Rightarrow\frac{\text{a}}{\text{x}}=\frac{3\text{k}}{2\text{m}}=\omega^2$
$\Rightarrow\omega=\sqrt{\frac{3\text{k}}{2\text{m}}}$
$\therefore$ time period $\text{T}=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{2\text{m}}{3\text{k}}}$
Because net force $\text{A }\&\text{ B}=\sqrt{\Big(\frac{\text{kx}}{2}\Big)^2+\Big(\frac{\text{kx}}{2}\Big)^2+2\Big(\frac{\text{kx}}{2}\Big)\Big(\frac{\text{kx}}{2}\Big)\cos120^\circ}=\frac{\text{kx}}{2}$$\therefore\text{a}=\frac{\text{F}}{\text{m}}=\frac{3\text{kx}}{2\text{m}}$
$\Rightarrow\frac{\text{a}}{\text{x}}=\frac{3\text{k}}{2\text{m}}=\omega^2$
$\Rightarrow\omega=\sqrt{\frac{3\text{k}}{2\text{m}}}$
$\therefore$ time period $\text{T}=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{2\text{m}}{3\text{k}}}$
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