A mass of 2.0 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 N/m. What should be the minimum amplitude of the motion, so that the mass gets detached from the pan? [AIPMT (S) 2007] |
(Take \[g=10m/{{s}^{2}}\]) |
A) 8.0 cm
B) 10.0 cm
C) Any value less than 12.0 cm
D) 4.0 cm
Correct Answer: B
Solution :
Let the minimum amplitude of SHM is a. Restoring force on spring |
\[F=ka\] |
Restoring force is balanced by weight mg of block. For mass to execute simple harmonic motion of amplitude a. |
\[\therefore \] \[ka=mg\] |
or \[a=\frac{mg}{k}\] |
Here, \[m=2\text{ }kg,\text{ }fc=200\text{ }N/m,\text{ }g=10\text{ }m/{{s}^{2}}\] |
\[\therefore \] \[a=\frac{2\times 10}{200}=\frac{10}{100}m\] |
\[=\frac{10}{100}\times 100\,cm=10\,cm\] |
Hence, minimum amplitude of the motion should be 10 cm, so that the mass gets detached from the pan. |
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