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(C) When a block of mass $M$ hangs from a spring in equilibrium,the downward gravitational force is balanced by the upward spring force.$Mg = kx$,wher

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$\frac{g}{\omega^2}$

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(C) When a block of mass $M$ hangs from a spring in equilibrium,the downward gravitational force is balanced by the upward spring force.$Mg = kx$,where $k$ is the spring constant and $x$ is the extension in the spring.Therefore,the extension $x = \frac{Mg}{k}$.We know that the angular frequency of a spring-mass system is given by $\omega = \sqrt{\frac{k}{M}}$.Squaring both sides,we get $\omega^2 = \frac{k}{M}$,which implies $k = M\omega^2$.Substituting the value of $k$ into the expression for $x$:$x = \frac{Mg}{M\omega^2} = \frac{g}{\omega^2}$.Thus,when the block is removed,the spring shortens by $\frac{g}{\omega^2}$.

$A$ block of mass $M$ hangs from a spring and oscillates vertically with an angular frequency $\omega$. If the block is removed from the spring,when it is in equilibrium position,the spring shortens by

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