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Simple Harmonic Motion — JEE physics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceJEE physicsSimple Harmonic Motion1 Mark
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 $2 m$ is suspended from it. The frequency of oscillation will now become
  • ✓
    $n$
  • B
    $2 n$
  • C
    $n / \sqrt{2}$
  • D
    $n(2)^{1 / 2}$
✓

Answer

Correct option: A.
$n$
$n=\frac{1}{2 \pi} \sqrt{\frac{k}{m}} \Rightarrow \frac{n}{n^{\prime}}=\sqrt{\frac{k}{m}} \times \frac{m^{\prime}{K^{\prime}}}=\sqrt{\frac{k}{m} \times \frac{2 m}{2 K}}=1 \Rightarrow n^{\prime}=n$

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