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A light, uniform, ideal spring is fixed at one end. If a mass is attached to the other end, the system oscillates with angular frequency $ω$ . Now the spring is fixed at both ends as shown in figure and is cut in half. The mass is attached between the two half springs. The new angular frequency of oscillations is

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$\frac{ω}{2}$

$2 ω$

$\sqrt{2} ω$

$ω$

answer is D.

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Detailed Solution

In case (1) $ω = \sqrt{\frac{k}{m}}$
In case (2) $ω^{1} = \sqrt{\frac{2 k + 2 k}{m}} = 2 ω$

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