Solution – Question 11
Solution: Box Stability in Oscillation
1. Condition for the Box to Remain Standstill
The box $M$ rests on the floor. It will lift off the floor (fail to remain standstill) if the normal reaction from the floor becomes zero.
Forces acting on the Box $M$:
- Weight $Mg$ (Downwards)
- Normal Reaction $N$ (Upwards)
- Spring Force exerted by the spring on the ceiling of the box.
If the spring is stretched, it pulls the ceiling down (increasing stability).
If the spring is compressed, it pushes the ceiling up.
Therefore, the danger of lifting occurs when the spring is in maximum compression (when mass $m$ is at its highest point).
2. Analyzing the Spring Force
Let $x_{eq}$ be the extension of the spring at equilibrium position for mass $m$.
$$ k x_{eq} = m g \implies x_{eq} = \frac{m g}{k} $$
During oscillation with amplitude $A$, the mass $m$ moves up and down by distance $A$ from the equilibrium position.
To achieve compression, the mass must move up past the natural length of the spring.
At the highest point (Maximum Compression):
Displacement from equilibrium = $A$ (upwards).
Net compression of spring ($C$) = $A – x_{eq}$ (assuming $A > x_{eq}$).
Upward force on the box ceiling ($F_s$) = $k(C) = k(A – x_{eq})$.
3. Solving for Maximum Amplitude
For the box to just lift off, the upward spring force must equal the weight of the box:
$$ F_s = Mg $$
$$ k(A – x_{eq}) = Mg $$
Substitute $x_{eq} = \frac{mg}{k}$:
$$ k \left( A – \frac{mg}{k} \right) = Mg $$
$$ kA – mg = Mg $$
$$ kA = Mg + mg $$
$$ A = \frac{(M + m)g}{k} $$
Answer: The maximum amplitude is $\frac{(M + m)g}{k}$.
