Q120 / 180 · AIIMS 2016 · Physics

A mass $M$ is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes simple harmonic oscillations with a time period $T$. If the mass is increased by $m$, then the time period becomes $\left(\frac{5}{4} T\right)$. The ratio of $\frac{m}{M}$ is

  • Correct answer
    A
    $9 / 16$
  • B
    $5 / 4$
  • C
    $25 / 16$
  • D
    $4 / 5$

Solution

$T=2 \pi \sqrt{\frac{M}{K}}$ and $T_1=2 \pi \sqrt{\frac{M+m}{K}}$ or $\frac{T_1}{T}=\sqrt{\frac{M+m}{M}}$ or $\frac{5}{4}=\left(1+\frac{m}{M}\right)^{\frac{1}{2}}$ or $\frac{m}{M}=\frac{25}{16}-1=\frac{9}{16}$ $\left(\because T_1=\frac{5}{4} T\right)$

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

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