Q165 / 200 · TS EAMCET 2001 · Physics

In order to double the frequency of the fundamental note emitted by a stretched string, the length is reduced to $\frac{3}{4}$ th of the original length and the tension is changed. The factor by which the tension is to be changed is

  • A
    $\frac{3}{8}$
  • B
    $\frac{2}{3}$
  • C
    $\frac{8}{9}$
  • Correct answer
    D
    $\frac{9}{4}$

Solution

Frequency of stretched string. $\mathrm{n}=\frac{1}{2 l} \frac{\sqrt{T}}{m}$ $\begin{aligned} & n \propto \frac{\sqrt{T}}{l} \\ & \frac{n_1}{n_2}=\frac{l_2}{I_1} \sqrt{\frac{T_1}{T_2}} \Rightarrow \frac{n}{2 n}=\frac{\frac{3}{4}}{l} \sqrt{\frac{T_1}{T_2}} \\ & \frac{1}{2}=\frac{3}{4} \sqrt{\frac{T_1}{T_2}} \Rightarrow \sqrt{\frac{T_1}{T_2}}=\frac{2}{3} \\ & \frac{T_1}{T_2}=\frac{4}{9} \Rightarrow T_2=\frac{9}{4} T_1\end{aligned}$ $\therefore$ Tension should be changed by the factor $\frac{9}{4}$.

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

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