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Simple Harmonic Motion — JEE physics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceJEE physicsSimple Harmonic Motion1 Mark
MCQ
A mass $M$ is suspended from a light spring. An additional mass $m$ added displaces the spring further by a distance $x$. Now the combined mass will oscillate on the spring with period
  • A
    $T=2 \pi \sqrt{(m g / x(M+m))}$
  • ✓
    $T=2 \pi \sqrt{((M+m) x / m g)}$
  • C
    $T=(\pi / 2) \sqrt{(m g / x(M+m))}$
  • D
    $T=2 \pi \sqrt{((M+m) / m g x)}$
✓

Answer

Correct option: B.
$T=2 \pi \sqrt{((M+m) x / m g)}$
As $m g$ produces extension $x$, hence $k=\frac{m g}{x}$
$\therefore T=2 \pi \sqrt{\frac{(M+m)}{k}}=2 \pi \sqrt{\frac{(M+m) x}{m g}}$

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