IF spring is in a already compressed state and two blocks are there one above another so when will the above block lose contact , what should be the condiition

1. Physical picture

• Two blocks of masses $m_1$ (top) and $m_2$ (bottom) are placed one above the other on a vertically compressed spring of force constant $k$.
• When the spring is released, it executes simple harmonic motion (SHM) with the combined mass $(m_1+m_2)$ acting like the load.
• Both blocks initially move together because of static friction/normal contact.

2. Contact criterion (Normal force)

• The normal force $N$ between the blocks is what keeps them in contact.

• Free–body diagram of the upper block while they still move together:

  • Upward forces: $N$
  • Downward forces: Weight $m_1 g$

• Using Newton’s second law (taking upward positive):
  $N - m_1 g = m_1 a$ where $a$ is the upward acceleration of the common motion at that instant.

• The instant the block is about to lose contact, $N \to 0$.

• Hence the condition becomes

  $0 - m_1 g = m_1 a \;\;\Rightarrow\;\; a = -g$

  (The negative sign merely tells us that the required upward acceleration magnitude is $g$.)
  In magnitude:

  $|a| \ge g$

3. Relating acceleration to spring parameters

For SHM of angular frequency $\omega$:

$\omega = \sqrt{\frac{k}{m_1 + m_2}}$

The displacement of the platform (lower block) may be written as

$x(t) = A \cos(\omega t + \phi)$

Acceleration is

$a(t) = -\omega^{2} A \cos(\omega t + \phi)$

Its maximum magnitude is

$a_{\text{max}} = \omega^{2} A$

Therefore, the separation happens if

$a_{\text{max}} \ge g\quad\Longrightarrow\quad A \ge \frac{g}{\omega^{2}} = g\,\frac{m_1+m_2}{k}$

So, either (i) the amplitude $A$ is big enough, or (ii) $k$ is large, or (iii) the masses are small, such that the spring can give at least $g$ upward acceleration.

Imagine this scenario like a pogo-stick!

1. Pogo-stick (spring) is already squeezed.
2. Two friends (blocks) are standing one on top of the other on that spring.
3. When the spring is released, it pushes both friends upward.
4. The top friend will only stay touching the bottom friend as long as the bottom friend can keep pushing him up faster than gravity pulls him down.
5. The moment gravity wins (that is, the spring cannot accelerate the bottom friend upward as fast as gravity pulls the top friend downward), the top friend will start to float — he loses contact!

So, in very simple words:

The top block loses contact when the spring’s upward jerk (acceleration) becomes equal to or more than the pull of gravity ($g$).

If the spring can give an upward acceleration $a$ to the system and at some instant $a \ge g$, the normal force between the blocks becomes zero, and the upper block separates.

Step-by-step derivation

1. System description

   • Upper mass: $m_1$
   • Lower mass: $m_2$
   • Spring constant: $k$
   • Amplitude of oscillation after release: $A$
   • Angular frequency:

   $\omega = \sqrt{\frac{k}{m_1 + m_2}}$

2. Equation for upper block while in contact

   $N - m_1 g = m_1 a$

3. Contact loss condition
   Set $N=0$:

   $-m_1 g = m_1 a \;\;\Longrightarrow\;\; a = -g$

   (Magnitude $a_{\text{req}} = g$ upward.)

4. Maximum acceleration provided by the spring

   $a_{\text{max}} = \omega^{2} A$

5. Condition to satisfy

   $\omega^{2} A \ge g$

   Substituting $\omega$:

   $\frac{k}{m_1 + m_2}\; A \; \ge \; g$

   $A \ge \frac{g\,(m_1 + m_2)}{k}$

6. Final Answer
   The upper block loses contact if the amplitude of oscillation is at least

   $\boxed{\displaystyle A \ge \frac{g\,(m_1 + m_2)}{k}}$

   equivalently, whenever the upward acceleration of the platform equals or exceeds $g$ so that the normal reaction drops to zero.