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In the arrangement shown in figure the sleeve $M$ of mass $m = 0.20 k g$ is fixed between two identical springs whose combined stiffness is equal to $k = 20 N / m$ . The sleeve can slide without friction over a horizontal bar $A B$ . The arrangement rotates with a constant angular velocity $ω = 4.4 r a d / s$ .about a vertical axis passing through the middle of the bar. The period of small oscillations of the sleeve is T and the minimum value of angular velocity for which will there be no oscillations of the sleeve is $ω_{0 .} T h e n$

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$ω = \sqrt{k / m} = 10 r a d / s$

$T = 2 π / \sqrt{\frac{k}{m} - ω^{2}} = 0.7 s$

$ω = \sqrt{4 k / m} = 20 r a d / s$

$T = 2 π / \sqrt{\frac{k}{2 m} - ω^{2}} = 1.5 s$

answer is A, C.

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Detailed Solution

Suppose the block is displaced slightly $(x)$ from its mean position, then restoring force

$F = - [s p r i n g f o r c e - c e n t r i f u g a l f o r c e] = - [k x - m ω^{2} x]$

Acceleration of the block

$a = \frac{F}{m} = (\frac{k}{m} - ω^{2}) (- x) ∴ T = \frac{2 π}{\sqrt{(\frac{k}{m} - ω^{2})}}$

For no oscillations $k x - m ω^{2} x \leq 0 o r ω \geq \sqrt{k / m}$

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