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The frequency of oscillations of a mass $m$ connected horizontally by a spring of spring constant $k$ is $4 \mathrm{~Hz}$. When the spring is replaced by two identical spring as shown in figure. Then the effective frequency is,
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The correct answer is:
$2 \sqrt{2}$
If one spring is connected then frequency $=4 \mathrm{~Hz}$
If two springs are connected in series, then frequency,
$v=\frac{1}{2 \pi} \sqrt{\frac{k^{\prime}}{m}}$
from eq ${ }^{\mathrm{n}}$ (i) and (ii)
$\begin{aligned} & \frac{4}{v}=\frac{1}{2 \pi} \sqrt{\frac{k}{m}} / \frac{1}{2 \pi} \sqrt{\frac{k}{2 m}} \\ & \frac{4}{v}=\sqrt{2} \text { or } \quad v=2 \sqrt{2} \mathrm{~Hz}\end{aligned}$
If two springs are connected in series, then frequency,
$v=\frac{1}{2 \pi} \sqrt{\frac{k^{\prime}}{m}}$
from eq ${ }^{\mathrm{n}}$ (i) and (ii)
$\begin{aligned} & \frac{4}{v}=\frac{1}{2 \pi} \sqrt{\frac{k}{m}} / \frac{1}{2 \pi} \sqrt{\frac{k}{2 m}} \\ & \frac{4}{v}=\sqrt{2} \text { or } \quad v=2 \sqrt{2} \mathrm{~Hz}\end{aligned}$
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