A ring of mass m is attached to a horizontal | vedclass

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Question

(A) Let the vertical distance from the rod to the fixed end of the spring be $h$. From the geometry at position $B$,we have $\cos 53^{\circ} = \frac{h

Vedclass · Accepted Answer

$\frac{2l_0}{3}\sqrt{\frac{k}{m}}$

Vedclass · Answer

(A) Let the vertical distance from the rod to the fixed end of the spring be $h$. From the geometry at position $B$,we have $\cos 53^{\circ} = \frac{h}{l_0+x}$ and $h = l_0$. Thus,$\frac{3}{5} = \frac{l_0}{l_0+x}$. Solving for $x$,we get $3(l_0+x) = 5l_0$,which implies $3x = 2l_0$,or $x = \frac{2}{3}l_0$. By the law of conservation of mechanical energy,the potential energy stored in the spring at position $B$ is converted into the kinetic energy of the ring at position $A$ (where the spring length is $l_0$ and potential energy is zero). $\frac{1}{2}kx^2 = \frac{1}{2}mv^2$. $v = x\sqrt{\frac{k}{m}} = \left(\frac{2}{3}l_0\right)\sqrt{\frac{k}{m}}$.

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