Question

Consider spring block pendulum system hanging in equilibrium. A bullet of mass m 2 moving at a speed u hits the block of mass m from downward direction and gets embedded in it as shown in Figure.

Answer 1

The amplitude of oscillation is $A = m_2 \sqrt{\frac{u^2}{k(m + m_2)} + \frac{g^2}{k^2}}$.

Answer 2

The problem involves a sequence of events: initial equilibrium, an inelastic collision, and subsequent simple harmonic motion (SHM).

Step 1: Initial Equilibrium of the Block

Before the collision, the block of mass $m$ is hanging in equilibrium. This means the upward spring force balances the downward gravitational force.
Let $x_0$ be the initial extension of the spring from its natural length. $k x_0 = m g$
So, $x_0 = \frac{m g}{k}$

Step 2: Inelastic Collision

A bullet of mass $m_2$ moving with speed $u$ hits the block from below and gets embedded in it. This is a perfectly inelastic collision. During the collision, linear momentum is conserved. The collision happens very quickly, so the position of the block does not change significantly during this instant.
Let $V$ be the velocity of the combined system (block + bullet) immediately after the collision.
By conservation of linear momentum in the vertical direction:
Initial momentum = Final momentum
$m_2 u + m(0) = (m + m_2) V$
$V = \frac{m_2 u}{m + m_2}$

Step 3: New Equilibrium Position

After the collision, the total mass hanging from the spring is $(m + m_2)$. The new equilibrium position for this combined mass will be different.
Let $x_{eq}'$ be the new equilibrium extension of the spring from its natural length.
$k x_{eq}' = (m + m_2)g$
So, $x_{eq}' = \frac{(m + m_2)g}{k}$

Step 4: Simple Harmonic Motion (SHM) and Amplitude Calculation

Immediately after the collision, the combined mass $(m + m_2)$ is at position $x_0$ (from natural length) and has an upward velocity $V$. This system will now oscillate about its new equilibrium position $x_{eq}'$.
To find the amplitude $A$ of the oscillation, we use the conservation of mechanical energy for SHM. The total energy of an oscillating system is the sum of its kinetic energy and potential energy (elastic potential energy due to displacement from the equilibrium position).

The displacement of the combined mass from its new equilibrium position at the instant of collision is:
$\Delta x = x_0 - x_{eq}'$
$\Delta x = \frac{m g}{k} - \frac{(m + m_2)g}{k}$
$\Delta x = \frac{m g - m g - m_2 g}{k}$
$\Delta x = -\frac{m_2 g}{k}$
The negative sign indicates that the position $x_0$ (where the collision occurred) is above the new equilibrium position $x_{eq}'$.

The total mechanical energy $E$ of the oscillating system can be expressed as:
$E = \frac{1}{2} (m + m_2)v^2 + \frac{1}{2} k (\text{displacement from new equilibrium})^2$

At the instant immediately after the collision, the velocity is $V$ and the displacement from the new equilibrium is $\Delta x$:
$E = \frac{1}{2} (m + m_2)V^2 + \frac{1}{2} k (\Delta x)^2$

At the maximum extension (or compression), which is the amplitude $A$, the velocity of the mass is momentarily zero:
$E = \frac{1}{2} k A^2$

Equating the two expressions for total energy:
$\frac{1}{2} k A^2 = \frac{1}{2} (m + m_2)V^2 + \frac{1}{2} k (\Delta x)^2$
Multiplying by 2:
$k A^2 = (m + m_2)V^2 + k (\Delta x)^2$

Now, substitute the expressions for $V$ and $\Delta x$:
$k A^2 = (m + m_2)\left(\frac{m_2 u}{m + m_2}\right)^2 + k\left(-\frac{m_2 g}{k}\right)^2$
$k A^2 = (m + m_2)\frac{m_2^2 u^2}{(m + m_2)^2} + k\frac{m_2^2 g^2}{k^2}$
$k A^2 = \frac{m_2^2 u^2}{m + m_2} + \frac{m_2^2 g^2}{k}$

Divide by $k$ to solve for $A^2$:
$A^2 = \frac{m_2^2 u^2}{k(m + m_2)} + \frac{m_2^2 g^2}{k^2}$

Taking the square root to find the amplitude $A$:
$A = \sqrt{\frac{m_2^2 u^2}{k(m + m_2)} + \frac{m_2^2 g^2}{k^2}}$
This can also be written as:
$A = m_2 \sqrt{\frac{u^2}{k(m + m_2)} + \frac{g^2}{k^2}}$

This $A$ represents the amplitude of oscillation, which is the maximum displacement from the new equilibrium position. In the context of such problems, "maximum extension" often refers to this amplitude.