Q132 / 160 · TS EAMCET 2010 · Physics

A sonometer wire has a length of $114 \mathrm{~cm}$, between two fixed ends. Where should two bridges be placed so as to divide the wire into three segments (in $\mathrm{cm}$ ) whose fundamental frequencies are in the ratio $1: 3: 4$ ?

  • A
    $l_1, l_2, l_3=18,24,72$
  • B
    $l_1, l_2, l_3=24,18,72$
  • C
    $l_1, l_2, l_3=72,18,24$
  • Correct answer
    D
    $l_1, l_2, l_3=72,24,18$

Solution

If length of the wire between the two bridges is $l$, the frequency of vibration $n=\frac{1}{2 l} \sqrt{\frac{T}{m}}$ $\therefore \quad n \propto \frac{1}{l}$ Therefore, $n_1: n_2: n_3=\frac{1}{l_1}: \frac{1}{l_2}: \frac{1}{l_3}$ $n_1: n_2: n_3=\frac{1}{\frac{114 \times 1}{8}}: \frac{1}{\frac{114 \times 3}{8}}: \frac{1}{\frac{114 \times 4}{8}}$ or $\quad n_1: n_2: n_3=8: \frac{8}{3}: 2$ or $\quad n_1: n_2: n_3=72: 24: 18$

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

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