Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A mass of $2.0\text{ kg}$ 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 $200\text{ N/m}$ . What should be the minimum amplitude of the motion, so that the mass gets detached from the pan? (Take $g=10\text{ m/s}^2$ )

8.0 cm

10.0 cm

any value less than 12.0 cm

4.0 cm

Step-by-Step Solution

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.
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 $a_{max} = \omega^2 A$ .
Set Condition: For detachment, $a_{max} \ge g$ . The minimum amplitude occurs when $a_{max} = g$ . $\omega^2 A = g$
Substitute $\omega$ : We know $\omega^2 = \frac{k}{m}$ . Therefore: $\frac{k}{m} A = g$ $A = \frac{mg}{k}$
Calculate Value: Given $m = 2.0\text{ kg}$ , $g = 10\text{ m/s}^2$ , and $k = 200\text{ N/m}$ . $A = \frac{2.0 \times 10}{200} = \frac{20}{200} = 0.1\text{ m}$ $A = 10.0\text{ cm}$

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