Question

A rod is fixed between a vertical wall and a horizontal surface. A smooth ring of mass 1 kg is released from rest which can move along the rod as shown. At the release point spring is vertical and relaxed. The natural length of the spring is $(\sqrt{3} + 1)$ m. Rod makes an angle of $30^\circ$ with the horizontal. Ring again comes to rest when spring makes an angle of $30^\circ$ with the vertical. (g = 10 m/s^2):

• A. Force constant of the spring is $\tfrac{5}{2}(\sqrt{3} + 1)$ N/m
• B. Maximum displacement of ring is $\tfrac{2}{\sqrt{3}-1}$ m
• C. Maximum extension in the spring is $(\sqrt{3} – 1)$ m
• D. Normal reaction on ring due to rod when it again comes to rest is $\tfrac{5}{2}(\sqrt{3}-1)$ N

Answer 1

Answer: (B)

Maximum displacement of ring is $\tfrac{2}{\sqrt{3}-1}$ m

Answer 2

Step 1. Geometry of final position
Let the natural length $L=\sqrt{3}+1$. If the ring moves a distance $s$ along the rod, its horizontal and vertical displacements satisfy

$$\tan30^\circ \;=\;\frac{s\cos30^\circ}{\,L + s\sin30^\circ\,} \;\Longrightarrow\;s=L.$$

Thus

$$s=\sqrt{3}+1 \;\equiv\;\frac{2}{\sqrt{3}-1}.$$

Step 2. Extension of spring
Final spring length $D$ makes $30^\circ$ with vertical so

$$D\cos30^\circ = L + s\sin30^\circ = L + \tfrac12L = \tfrac32L \;\Longrightarrow\; D = \frac{3L}{\sqrt3}=L\sqrt3.$$

Extension $x=D-L = L(\sqrt3-1)=2$ m.

Step 3. Force‐balance along the rod
At final rest, component of weight along rod $=mg\sin30^\circ=5$ N balances the spring force component:

$$k\,x\,\frac{(s+\tfrac12L)}{D}=5 \;\Longrightarrow\; k\;\text{cancels out to give}\;s=L\;\text{and no change in }s.$$

Conclusion. Maximum displacement $s = \frac{2}{\sqrt3-1}$ m.