Question

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. 10 . 0 cm
• B. any value less than 12.0 cm
• C. 4. 0 cm
• D. 8.0 cm

Answer 1

Answer: (A)

Consider the minimum amplitude of SHM is a As we know that, the restoring force on spring is
$F = Ka$
Restoring force is balanced by weight mg of block and for mass to execute SHM of amplitude a.
Therefore, $Ka = mg$
$\Rightarrow a =\frac{ mg }{ K }$
$ m =2 \,Kg $
$ K =200 \,N / m $
$g=10 \, m / S ^{2}$
Therefore,
$a =\frac{2 \times 10}{200} $
$=\frac{10}{100} \, m$
$=\frac{10}{100} \times 100 \, cm =10 \, cm$
So, the minimum amplitude of the motion should be $10\,cm$, so that the mass get detached from the pan.