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A mass of 2.0 kg is put on a flat pan attached to a vertical spring fixed on the ground. 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
Detailed Solution
The mass can get detached from the pan only at the upper extreme position, where the acceleration is directed downward and has its maximum value $\omega^2 a$.
At that point, taking downward as positive: $mg - R = m\omega^2 a$, where $R$ is the normal reaction of the pan.
$R = mg - m\omega^2 a$
The mass just loses contact when $R = 0$, i.e. $g = \omega^2 a$
$\Rightarrow a = \dfrac{g}{\omega^2} = \dfrac{mg}{K}$, as $\omega^2 = \dfrac{K}{m}$
$\Rightarrow a = \dfrac{2 \times 10}{200}$
$\Rightarrow a = \dfrac{1}{10}$ m $= 10$ cm
So the minimum amplitude is 10.0 cm.
At that point, taking downward as positive: $mg - R = m\omega^2 a$, where $R$ is the normal reaction of the pan.
$R = mg - m\omega^2 a$
The mass just loses contact when $R = 0$, i.e. $g = \omega^2 a$
$\Rightarrow a = \dfrac{g}{\omega^2} = \dfrac{mg}{K}$, as $\omega^2 = \dfrac{K}{m}$
$\Rightarrow a = \dfrac{2 \times 10}{200}$
$\Rightarrow a = \dfrac{1}{10}$ m $= 10$ cm
So the minimum amplitude is 10.0 cm.