A block of mass m attached to a massless spri | vedclass

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(B) At the equilibrium position,the velocity of the block is maximum,given by $V_0 = \omega_0 A = \sqrt{\frac{k}{m}} A$.When half the mass breaks off

Vedclass · Accepted Answer

$1/\sqrt{2}$

Vedclass · Answer

(B) At the equilibrium position,the velocity of the block is maximum,given by $V_0 = \omega_0 A = \sqrt{\frac{k}{m}} A$.When half the mass breaks off at the equilibrium position,the velocity of the remaining mass $m/2$ remains the same as $V_0$ because there is no impulsive force acting on the system in the horizontal direction.Let the new amplitude be $A'$. The new angular frequency is $\omega' = \sqrt{\frac{k}{m/2}} = \sqrt{\frac{2k}{m}} = \sqrt{2} \omega_0$.Since the velocity at the equilibrium position is $V_0 = \omega' A'$,we have:$\omega_0 A = \omega' A'$$\omega_0 A = (\sqrt{2} \omega_0) A'$$A' = \frac{A}{\sqrt{2}}$.Thus,$f = \frac{1}{\sqrt{2}}$.

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