A mass at the end of a spring executes harmon | vedclass

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Question

(A) For a mass-spring system executing simple harmonic motion,the maximum speed $V$ at the equilibrium position is given by the formula $V = A \omega$

Vedclass · Accepted Answer

$2V$

Vedclass · Answer

(A) For a mass-spring system executing simple harmonic motion,the maximum speed $V$ at the equilibrium position is given by the formula $V = A \omega$,where $A$ is the amplitude and $\omega$ is the angular frequency.Since the angular frequency $\omega = \sqrt{\frac{k}{m}}$ depends only on the spring constant $k$ and the mass $m$,it remains constant regardless of the amplitude.Given the initial amplitude is $A$,the initial speed is $V = A \omega$.When the amplitude is increased to $A' = 2A$,the new maximum speed $V'$ at the equilibrium position becomes $V' = A' \omega = (2A) \omega = 2(A \omega) = 2V$.Therefore,the speed of the mass passing through the equilibrium position will be $2V$.

$A$ mass at the end of a spring executes harmonic motion about an equilibrium position with an amplitude $A$. Its speed as it passes through the equilibrium position is $V$. If extended $2A$ and released,the speed of the mass passing through the equilibrium position will be

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