JCECE Engineering JCECE Engineering Solved Paper-2007

  • question_answer
    Two identical springs of spring constant \[k\] are connected in series and parallel as shown in figure. A mass \[M\] is suspended from them. The ratio of their frequencies of vertical oscillation will be

    A) \[2:1\]                                  

    B) \[1:2\]

    C) \[4:1\]                                  

    D)  \[1:4\]

    Correct Answer: B

    Solution :

    Key Idea: In first set up, the springs are joined in series and in second, the springs are joined in parallel. When springs are connected in series the effective spring constant is                 \[\frac{1}{k}=\frac{1}{k}+\frac{1}{k}=\frac{2}{k}\] \[\Rightarrow \]               \[k'=\frac{k}{2}\] Hence, frequency\[n'=\frac{1}{2\pi }\sqrt{\frac{k'}{m}}\]                 \[=\frac{1}{2\pi }\sqrt{\left( \frac{k}{2m} \right)}\]                                          ? (i) When springs are connected in parallel, the effective force constant is                 \[k'\,\,'=k+k=2k\] Therefore, frequency is                 \[n'\,\,'=\frac{1}{2\pi }\sqrt{\frac{2k}{m}}\]                                         ? (ii) Dividing Eq. (i) by Eq. (ii), we get                 \[\frac{n'}{n'\,\,'}=\frac{\sqrt{(k/2m)}}{\sqrt{(2k/m)}}=\frac{1}{2}\]


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