A mass of $2.0 ext{ kg}$ is put on a flat pan attached to a vertical spring fixed on the ground as shown in t

Question

A mass of $2.0	ext{ kg}$ is put on a flat pan attached to a vertical spring fixed on the ground as shown in the figure. The mass of the spring and the pan is negligible. When pressed slightly and released, the mass executes a simple harmonic motion. The spring constant is $200	ext{ N/m}$. What should be the minimum amplitude of the motion, so that the mass gets detached from the pan? (Take $g=10	ext{ m/s}^2$)

Accepted Answer

1. **Identify Condition for Detachment:** The mass will detach from the pan when the contact force (Normal Reaction, $N$) between the mass and the pan becomes zero. This happens when the downward acceleration of the pan equals the acceleration due to gravity ($g$). If the pan accelerates downwards faster than $g$, the mass (which falls at $g$ under gravity alone) will lose contact.
2. **Analyze Motion:** In a vertical spring-mass system, the maximum downward acceleration occurs at the highest point of the oscillation (extreme position). The magnitude of maximum acceleration in SHM is given by $a_{max} = \omega^2 A$ .
3. **Set Condition:** For detachment, $a_{max} \ge g$. The minimum amplitude occurs when $a_{max} = g$.
   $$ \omega^2 A = g $$
4. **Substitute $\omega$:** We know $\omega^2 = \frac{k}{m}$ . Therefore:
   $$ \frac{k}{m} A = g $$
   $$ A = \frac{mg}{k} $$
5. **Calculate Value:** Given $m = 2.0	ext{ kg}$, $g = 10	ext{ m/s}^2$, and $k = 200	ext{ N/m}$.
   $$ A = \frac{2.0 	imes 10}{200} = \frac{20}{200} = 0.1	ext{ m} $$
   $$ A = 10.0	ext{ cm} $$

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