A spring has a natural length l with one end | vedclass

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(B) Let the vertical distance be $l$. The length of the spring at any angle $	heta$ with the vertical is $h = l / \cos 	heta$. The extension in the

Vedclass · Accepted Answer

$\cos ^{-1}\left(\frac{2}{2+\sqrt{3}}\right)$

Vedclass · Answer

(B) Let the vertical distance be $l$. The length of the spring at any angle $	heta$ with the vertical is $h = l / \cos 	heta$. The extension in the spring is $x = h - l = l(1/\cos 	heta - 1)$.By conservation of energy,the potential energy stored in the spring is converted into kinetic energy. The maximum velocity $v_{max}$ occurs when the spring is at its natural length (i.e.,$	heta = 0$,$x = 0$),but here the ring is constrained to the rod. The potential energy at the initial position $(	heta = 60^{\circ})$ is $U_i = \frac{1}{2} k x_i^2$,where $x_i = l(1/\cos 60^{\circ} - 1) = l(2-1) = l$. So $U_i = \frac{1}{2} k l^2$.At any angle $	heta$,the energy equation is $\frac{1}{2} k l^2 = \frac{1}{2} m v^2 + \frac{1}{2} k (l/\cos 	heta - 1)^2 l^2$. The maximum velocity $v_{max}$ occurs when the extension is zero,i.e.,at $	heta = 0$. However,the ring is constrained to the rod,so the minimum extension is at $	heta = 0$,where $x=0$. Thus,$\frac{1}{2} m v_{max}^2 = \frac{1}{2} k l^2$. Given $v = \frac{1}{2} v_{max}$,then $v^2 = \frac{1}{4} v_{max}^2$. Substituting this into the energy equation: $\frac{1}{2} k l^2 = \frac{1}{2} m (\frac{1}{4} v_{max}^2) + \frac{1}{2} k l^2 (1/\cos 	heta - 1)^2$. Since $\frac{1}{2} m v_{max}^2 = \frac{1}{2} k l^2$,we have $\frac{1}{2} k l^2 = \frac{1}{8} k l^2 + \frac{1}{2} k l^2 (1/\cos 	heta - 1)^2$. Dividing by $\frac{1}{2} k l^2$: $1 = 1/4 + (1/\cos 	heta - 1)^2 \Rightarrow (1/\cos 	heta - 1)^2 = 3/4 \Rightarrow 1/\cos 	heta - 1 = \sqrt{3}/2 \Rightarrow 1/\cos 	heta = 1 + \sqrt{3}/2 = (2+\sqrt{3})/2$. Thus,$\cos 	heta = 2 / (2+\sqrt{3})$,so $	heta = \cos^{-1}(2 / (2+\sqrt{3}))$. The correct option is $B$.

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