BCECE Medical BCECE Medical Solved Papers-2005

  • question_answer
    The fundamental frequency of a sonometer wire is n. If its radius is doubled and its tension becomes half, the material of the wire remains same, the new fundamental frequency will be:

    A)  n             

    B)  \[\frac{n}{\sqrt{2}}\]

    C)  \[\frac{n}{2}\]

    D)  \[\frac{n}{2\sqrt{2}}\]

    Correct Answer: D

    Solution :

    Frequency of sonometer wire is given by                 \[n=\frac{1}{2}\sqrt{\frac{T}{m}}\] where m is mass of string per unit length, and T is tension in the string. Also,      \[m=\pi {{r}^{2}}d\] r being radius of string and d is the density of material of string. So,          \[n=\frac{1}{2\,l}\sqrt{\frac{T}{\pi {{r}^{2}}d}}\] or            \[n\propto \frac{\sqrt{T}}{r}\] or            \[\frac{{{n}_{1}}}{{{n}_{2}}}=\sqrt{\frac{{{T}_{1}}}{{{T}_{2}}}}\times \left( \frac{{{r}_{2}}}{{{r}_{1}}} \right)\] Given,   \[{{r}_{2}}=2\,{{r}_{1}},\,\,\,\,{{T}_{2}}=\frac{{{T}_{1}}}{2},\,\,\,\,{{n}_{1}}=n\] Hence,  \[\frac{n}{{{n}_{2}}}=\sqrt{2}\times 2\] or \[{{n}_{2}}=\frac{n}{2\sqrt{2}}\]


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