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? (Take\[g=10\,m/{{s}^{2}}\])
A) 8.0 cm
B) 10.0cm
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, \[m=2kg,\,k=200\,N/m,\,g=10\,m/{{s}^{2}}\] \[\therefore \] \[a=\frac{2\times 10}{200}=\frac{10}{100}m\] \[=\frac{10}{100}\times 100\,cm=10\,cm\] Hence, minimum amplitude of the motion should be 10 cm, so that the mass gets detached from the pan.
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