Q.A body of mass $m$ is attached to one end of a massless spring which is suspended vertically from a fixed point. The mass is held in hand so that the spring is neither stretched nor compressed. Suddenly the support of the hand is removed. The lowest position attained by the mass during oscillation is 4cm below the point, where it was held in hand.

Concept: Simple Harmonic Motion Energy — the mass falls, converting gravitational potential into spring potential, then oscillates about the equilibrium point.

Reasoning:

1. When released from the unstretched position, the mass falls to a lowest point 4 cm below. At that lowest point, the spring is stretched by 4 cm and the mass momentarily stops. The equilibrium position (where net force is zero) lies halfway between the top and bottom of the motion, because the spring force is linear.

2. For a vertical spring, the equilibrium extension is $mg/k$. The amplitude $A$ is the distance from equilibrium to either extreme. Since the top extreme is at zero stretch and the bottom extreme is at 4 cm stretch, the equilibrium is at 2 cm stretch. Hence $A = 2\ \text{cm}$. …

When you release a mass from the unstretched position of a vertical spring, it doesn't just fall — it oscillates. The key insight is that the motion is simple harmonic about the equilibrium point where the spring force balances gravity. The lowest point is not the equilibrium; it's the bottom of the oscillation, twice as far from the release point as the equilibrium is.

Let's work through this carefully.

1. Identify the equilibrium position.

   When the mass hangs at rest, the spring is stretched by an amount $x_0$ such that $k x_0 = mg$. This $x_0$ is the equilibrium stretch. The mass oscillates symmetrically about this point.

2. Relate the release point to the motion.

   The mass is released from the unstretched position — that is, from a point where the spring is neither stretched nor compressed. This is the top extreme of the oscillation. The mass then falls, passes through equilibrium, and reaches the bottom extreme at the lowest point. The distance from the top extreme to the bottom extreme is $2A$, where $A$ is the amplitude.

3. Use the given data.

   The lowest position is 4 cm below the release point. So the total vertical distance from top to bottom is 4 cm. Therefore:

That answers part (a).

4. Find the equilibrium stretch $x_0$. Since the top extreme is at the unstretched position and the equilibrium is $A$ below it, we have: $$$$

Step 1: The mass is released from the unstretched (natural-length) position with zero velocity — this is one extreme of the SHM, not the equilibrium.

Step 2: The equilibrium position is where the spring force balances gravity: $kx_0 = mg$, a distance $x_0$ below the natural-length point. By symmetry of SHM, the release point (top extreme) is a distance $A$ above equilibrium, so $x_0 = A$.

Step 3: The two extremes (top and bottom) are separated by $2A$. Since the lowest point is $4\ \text{cm}$ below the release point, $2A = 4\ \text{cm} \Rightarrow A = 2\ \text{cm} = 0.02\ \text{m}$. This also gives the equilibrium stretch $x_0 = 2\ \text{cm}$. …