Q146 / 160 · TS EAMCET 2011 · Physics

Two uniform stretched strings $A$ and $B$, made of steel, are vibrating under the same tension. If the first overtone of $A$ is equal to the second overtone of $B$ and if the radius of $A$ is twice that of $B$, the ratio of the lengths of the strings is

  • A
    2 : 3
  • B
    1 : 2
  • Correct answer
    C
    1 : 3
  • D
    1 : 4

Solution

Given, 1st overtone of $A=2$ nd overtone of $B$ $\begin{aligned} \Rightarrow & \frac{1}{l_1} \sqrt{\frac{T}{\pi r_1^2 d}} =\frac{3}{2 l_2} \sqrt{\frac{T}{\pi r_2^2 d}} \\ \Rightarrow & \frac{r_2}{r_1} =\frac{3}{2}\left(\frac{l_1}{l_2}\right) \end{aligned}$ But $r_1=2 r_2$ $\begin{aligned} & \therefore \quad \frac{r_2}{2 r_2}=\frac{3}{2}\left(\frac{l_1}{l_2}\right) \\ & \text { or } \quad \frac{l_1}{l_2}=\frac{1}{3} \Rightarrow l_1: l_2=1: 3 \\ & \end{aligned}$

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

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