NEET Physics Wave Mechanics NEET PYQ-Wave Mechanics

  • question_answer
    A sonometer wire when vibrated in full length has frequency n. Now it is divided by the help of bridges into a number of segments of lengths\[{{l}_{1}},\,{{l}_{2}},\,{{l}_{3}}...\]. When vibrated these segments have frequencies \[{{n}_{1}},\,{{n}_{2}},\,{{n}_{3}},....\] then the correct relation is:                                           [AIPMT 2000]

    A)  \[n={{n}_{1}}+{{n}_{2}}+{{n}_{3}}+...\]

    B)       \[{{n}^{2}}=n_{1}^{2}+n_{2}^{2}+n_{3}^{2}+...\]

    C)  \[\frac{1}{n}=\frac{1}{{{n}_{1}}}+\frac{1}{{{n}_{2}}}+\frac{1}{{{n}_{3}}}+...\]

    D)              \[\frac{1}{\sqrt{n}}=\frac{1}{\sqrt{{{n}_{1}}}}+\frac{1}{\sqrt{{{n}_{2}}}}+\frac{1}{\sqrt{{{n}_{3}}}}+...\]

    Correct Answer: C

    Solution :

    From law of length, the frequency of vibrating string is inversely proportional to its length, i.e.,
                            \[n\propto \frac{1}{l}\]
    or         \[nl=\]constant (say k)
    or         \[nl=k\]
    or         \[l=\frac{k}{n}\]
                The segment of string of length \[{{l}_{1}},\,{{l}_{2}},\,{{l}_{3}}...\] have frequencies \[{{n}_{1}},{{n}_{2}},{{n}_{3}}\,,....\]
                Total length of string is \[l\].
                So,       \[l={{l}_{1}}+{{l}_{2}}+{{l}_{3}}+...\]
                \[\therefore \]      \[\frac{k}{n}=\frac{k}{{{n}_{1}}}+\frac{k}{{{n}_{2}}}+\frac{k}{{{n}_{3}}}+.....\]
                or         \[\frac{1}{n}=\frac{1}{{{n}_{1}}}+\frac{1}{{{n}_{2}}}+\frac{1}{{{n}_{3}}}+....\]


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