Q116 / 120 · JEE Main 2007 · Physics

Two springs, of force constants $k_1$ and $k_2$, are connected to a mass $\mathrm{m}$ as shown. The frequency of oscillation of the mass is $\mathrm{f}$. If both $k_1$ and $k_2$ are made four times their original values, the frequency of oscillation becomes Figure

  • A
    $\mathrm{f / 2}$
  • B
    $\mathrm{f / 4}$
  • C
    $4 \mathrm{f}$
  • Correct answer
    D
    $2 \mathrm{f}$

Solution

$\begin{aligned} f & =\frac{1}{2 \pi} \sqrt{\frac{k_1+k_2}{m}} \\ f^{\prime} & =\frac{1}{2 \pi} 2 \sqrt{\frac{k_1+k_2}{m}}=2 f\end{aligned}$

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

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