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  • A mass  is put on a flat pan attached to a vertical spring fixed on the ground. The mass of the spring

A mass m=1.0kg 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 simple harmonic motion. The spring constant is 500\: N/m. What is the amplitude A of the motion, so that the mass m tends to get detached from the pan ?  (Take g=10m/s^{2}). The spring is stiff enough so that it does not get disorted during the motion. 

Option: 1

A> 2.0\: cm
 


Option: 2

A= 2.0\: cm


Option: 3

A< 2.0\: cm

 


Option: 4

A= 1.5\: cm


Answers (1)

best_answer

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{m g}{K}$ \\ Here, we have $$ m=2 K g $$ \\ $K=200 N / m$ \\ $g=10 m / S^{2}$ \\ Therefore, \\ \\ $\begin{aligned} a &=\frac{2 \times 10}{10^{200}} \\ &=\frac{10}{100} m \\ &=\frac{10}{100} \times 100 cm =10 cm \end{aligned}

So, the minimum amplitude of the motion should be 10cm, so that the mass get detached from the pan.

Posted by

Ajit Kumar Dubey

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