A) 1:2
B) 2:1
C) 1:3
D) 3:1
Correct Answer: A
Solution :
Key Idea: In first arrangement springs are in parallel and in second arrangement springs are in series. In first case, springs are connected in parallel, so their equivalent spring constant \[{{k}_{p}}={{k}_{1}}+{{k}_{2}}\] So, frequency of this spring block system is given by \[{{f}_{p}}=\frac{1}{2\,\pi }\sqrt{\frac{{{k}_{p}}}{m}}\] or \[{{f}_{p}}=\frac{1}{2\pi }\sqrt{\frac{{{k}_{1}}+{{k}_{2}}}{m}}\] but \[{{k}_{1}}={{k}_{2}}\] \[\therefore \] \[{{f}_{p}}=\frac{1}{2\pi }\sqrt{\frac{2\,k}{m}}\] ?.. (i) Now, in second case, springs are connected in series, so their equivalent spring constant \[k=\frac{{{k}_{1}}{{k}_{2}}}{{{k}_{1}}+{{k}_{2}}}\] Hence, frequency of this arrangement is given by \[{{f}_{s}}=\frac{1}{2\pi }\sqrt{\frac{{{k}_{1}}{{k}_{2}}}{({{k}_{1}}+{{k}_{2}})}}\] or \[{{f}_{s}}=\frac{1}{2\pi }\sqrt{\frac{k}{2m}}\] ?? (ii) Dividing Eq. (ii) by Eq. (i), we get \[\frac{{{f}_{s}}}{{{f}_{p}}}\frac{\frac{1}{2\pi }\sqrt{\frac{k}{2m}}}{\frac{1}{2\pi }\sqrt{\frac{2k}{m}}}=\sqrt{\frac{1}{4}}\] or \[\frac{{{f}_{s}}}{{{f}_{p}}}=\frac{1}{2}\]
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