MCQ
A weightless spring which has a force constant oscillates with frequency $n$ when a mass $m$ is suspended from it. The spring is cut into two equal halves and a mass $2m $ is suspended from it. The frequency of oscillation will now become
  • $n$
  • B
    $2n$
  • C
    $\frac{n}{\sqrt2}$
  • D
    $n(2)^{1/2}$

Answer

Correct option: A.
$n$
a
(a) $n = \frac{1}{{2\pi }}\sqrt {\frac{k}{m}} $

==>$\frac{n}{{n'}} = \sqrt {\frac{k}{m} \times \frac{{m'}}{{K'}}} $

$ = \sqrt {\frac{k}{m} \times \frac{{2m}}{{2K}}} = 1$

==> $n' = n$

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