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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $B$.

Similar Questions

The potential energy of a long spring when stretched by $2 \ cm$ is $U$. If the spring is stretched by $8 \ cm$,the potential energy stored in it is:

$A$ $1\, kg$ block moves towards a light spring with a velocity of $8\, m/s$. When the spring is compressed by $3\, m$,its momentum becomes half of the original momentum. The spring constant of the spring is:

$A$ ball of mass $m$ is dropped from a height $h$ on a platform fixed at the top of a vertical spring,as shown in the figure. The platform is depressed by a distance $x$. Then the spring constant is

Vedclass Test Series

Exam Paper Generator

Online Exam Module