Q142 / 160 · AP EAMCET 2007 · Physics

A segment of wire vibrates with a fundamental frequency of $450 \mathrm{~Hz}$ under a tension of $9 \mathrm{~kg} \mathrm{wt}$. Then tension at which the fundamental frequency of the same wire becomes $900 \mathrm{~Hz}$ is

  • Correct answer
    A
    $36 \mathrm{~kg}-\mathrm{wt}$
  • B
    $27 \mathrm{~kg}-\mathrm{wt}$
  • C
    $18 \mathrm{~kg}-\mathrm{wt}$
  • D
    $72 \mathrm{~kg}-\mathrm{wt}$

Solution

Fundamental frequency of wire $\begin{array}{rlrl}f & =\frac{1}{2 \pi} \sqrt{\frac{T}{m}} \\ \text { or } & f & \propto \sqrt{T} \\ \text { or } & \frac{f_2}{f_1} & =\sqrt{\frac{T_2}{T_1}} \\ \text { or } & \frac{900}{450} & =\sqrt{\frac{T_2}{9}} \\ \text { or } & T_2 & =4 \times 9=36 \mathrm{~kg}-\mathrm{wt}\end{array}$

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

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