A stiff spring having spring constant k = 400 | vedclass

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Question

(A) The angular frequency of oscillation is given by $\omega = \sqrt{\frac{k}{m}}$.For the block to lose contact with the spring, the upward accelerat

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(A) The angular frequency of oscillation is given by $\omega = \sqrt{\frac{k}{m}}$.For the block to lose contact with the spring, the upward acceleration of the block must exceed the acceleration due to gravity $g$.The block loses contact when the upward acceleration $a = \omega^2 x$ equals $g$, where $x$ is the displacement from the equilibrium position.At the equilibrium position, the spring is compressed by $x_0 = \frac{mg}{k} = \frac{10 	imes 10}{400} = 0.25 	ext{ m} = 25 	ext{ cm}$.When the block is at the highest point of its oscillation, the restoring force is directed downwards. The block loses contact when the upward acceleration equals $g$. Since the maximum upward acceleration in $SHM$ is $\omega^2 A$, where $A$ is the amplitude, the block will lose contact if $A > x_0$.The question asks for the extension (or rather, the displacement from the natural length) at which contact is lost. The block loses contact when the spring force becomes zero, which happens when the spring returns to its natural length ($x = 0$ relative to natural length).However, in the context of this standard problem, the block loses contact when the upward acceleration equals $g$. This occurs at the equilibrium position if the amplitude $A = x_0 = 25 	ext{ cm}$.

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