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

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Question

(A) $1$. In the equilibrium position,the spring is stretched by an amount $x_0 = M_1g/k$.$2$. When the block is pulled down by a distance $x$ and rele

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) $1$. In the equilibrium position,the spring is stretched by an amount $x_0 = M_1g/k$.$2$. When the block is pulled down by a distance $x$ and released,it performs simple harmonic motion with amplitude $A = x$.$3$. The maximum upward force exerted by the spring on the frame occurs when the block is at its highest point of oscillation (at a distance $x$ above the equilibrium position).$4$. At this highest point,the extension in the spring is $(x_0 - x)$. However,the force exerted by the spring on the top bar is upward,given by $F_s = k(x_0 + x)$ if we consider the extension from the natural length. More simply,the maximum upward force the spring exerts on the frame is $k(x_0 + x) = k(M_1g/k + x) = M_1g + kx$.$5$. For the frame to leave the floor,the upward force exerted by the spring on the frame must exceed the weight of the frame,$M_2g$.$6$. Thus,$M_1g + kx = M_2g$ is the condition for the frame to just lift off. However,the spring force at the top of the oscillation is $k(x - x_0)$ if $x > x_0$. The upward force on the frame is $k(x - x_0) = M_2g$.$7$. Substituting $x_0 = M_1g/k$,we get $k(x - M_1g/k) = M_2g$,which simplifies to $kx - M_1g = M_2g$.$8$. Therefore,$kx = (M_1 + M_2)g$,or $x = (M_1 + M_2)g/k$.

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