A block of mass M_1 is hung by a light spring | vedclass

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Question

(A) At equilibrium,the spring is stretched by $x_e = M_1g/k$.When the block is pulled down by a distance $x$ from the equilibrium position and release

Vedclass · Accepted Answer

$x = (M_1 + M_2)g/k$

Vedclass · Answer

(A) At equilibrium,the spring is stretched by $x_e = M_1g/k$.When the block is pulled down by a distance $x$ from the equilibrium position and released,it performs simple harmonic motion with amplitude $x$.The maximum upward force exerted by the spring on the frame occurs when the block is at its highest point (at a distance $x$ above the equilibrium position).At this highest point,the spring is compressed by $(x - x_e)$.The upward force exerted by the spring on the frame is $F_{up} = k(x - x_e)$.For the frame to leave the floor,the upward force must exceed the weight of the frame $M_2g$.Thus,$k(x - x_e) = M_2g$.Substituting $x_e = M_1g/k$,we get $k(x - M_1g/k) = M_2g$.$kx - M_1g = M_2g$.$kx = (M_1 + M_2)g$.$x = (M_1 + M_2)g/k$.

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