Q108 / 150 · MHT CET 2020 (19 Oct Shift 1) · Physics

When a mass ' $\mathrm{m}$ ' is suspended from a spring of length ' $\ell^{\prime}$, the length of the spring becomes ' $\mathrm{L}^{\prime}$. The mass is pulled down by a distance ' $\mathrm{d}$ ' and released. If the equation of motion of the mass is $\frac{d^{2} x}{d t^{2}}+\mathrm{P}^{2} x=0$, then $\mathrm{P}$ is equal to $(\mathrm{g}=$ acceleration due to gravity $)$

  • A
    $\frac{\mathrm{L}-\ell}{\mathrm{g}}$
  • B
    $\frac{\mathrm{g}}{\mathrm{L}-\ell}$
  • Correct answer
    C
    $\sqrt{\frac{\mathrm{g}}{\mathrm{L}-\ell}}$
  • D
    $\sqrt{\frac{\mathrm{L}-\ell}{\mathrm{g}}}$

Solution

INTIAL LENGTH $=l$ FINAL LENGTH=L. NET DISTANCE/LENGTH PULLED $=L-l$ $\therefore T I M E P E R I O D=2 \pi \sqrt{\frac{L-\ell}{g}}$ allo, $P=\frac{2 \pi}{T .}$ $\begin{aligned} \therefore P=& \frac{g}{L-l} . \\ &(\text { option }-3) . \end{aligned}$

Hence, the correct answer is option: 'C'.

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