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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Oscillations
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A spring-mass system is released from rest as shown in the figure . Initially the spring is unstretched . The equilibrium position of the system is at distance of from the initial state .$\;(g=10\;m/s^2)$

$(a)\;2\;m\qquad(b)\;4\;m\qquad(c)\;1\;m\qquad(d)\;3\;m$

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Answer : (a) $\;2\;m$
At equilibrium position , net force on the body is zero
Therefore , Kx-mg=0
Kx=mg
$10\times x=2 \times 10$
$x=2\;m\;.$
answered Mar 8, 2014 by yamini.v
 

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