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Q.

A block of mass m attached to a 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 fA. The value of f is

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a

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

b

$1$

c

$\frac{1}{2}$

d

$\sqrt{2}$

answer is D.

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Detailed Solution

Question Image

$V_{1} = V_{\max} = A ω A_{1} ω_{1} = A_{2} ω_{2}$

$A ω = A' ω' A \sqrt{\frac{k}{m}} = A_{2} \sqrt{\frac{K}{(m / 2)}}$

$A (\sqrt{k} / \sqrt{m}) = A' (\sqrt{2} k / \sqrt{m}) A = A_{2} \sqrt{2}$

$A' = A / \sqrt{2} A_{2} = \frac{A}{\sqrt{2}} = f A \to f = \frac{1}{\sqrt{2}}$

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A block of mass m attached to a 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 fA. The value of f is