A block of mass m attached to massless spring is

Question:

A block of mass $m$ attached to massless spring is performing oscillatory motion of amplitude 'A' on a frictionless horizontal plane. If half of the mass of the block breaks off when it is passing through its equilibrium point, the amplitude of oscillation for the remaining system become $f \mathrm{~A}$. The value of $f$ is:

  1. $\frac{1}{2}$

  2. $\sqrt{2}$

  3. 1

  4. $\frac{1}{\sqrt{2}}$


Correct Option: , 4

Solution:

At equilibrium position

$V_{0}=\omega_{0} A=\sqrt{\frac{K}{m}} A$......(i)

$V=\omega A^{\prime}=\sqrt{\frac{K}{\frac{m}{2}} A^{\prime}}$$\ldots . .(\mathrm{ii})$

$\therefore \quad A^{1}=\frac{A}{\sqrt{2}}$

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