A) 2k
B) k
C) \[\frac{k}{2}\]
D) 2048
Correct Answer: C
Solution :
Effective force constant is equal to the reciprocal of the sum of individual force constants, hence \[\frac{1}{{{k}_{e}}}=\frac{1}{{{k}_{1}}}+\frac{1}{{{k}_{2}}}+\frac{1}{{{k}_{3}}}+...\] Given, \[{{k}_{1}}=k,{{k}_{2}}=2k,{{k}_{3}}=4k,\] \[\therefore \] \[\frac{1}{{{k}_{e}}}=\frac{1}{k}+\frac{1}{2k}+\frac{1}{4k}+\frac{1}{8k}+....\] The given series is a geometric progression series, hence sum is \[{{S}_{x}}=\frac{a}{1-r}\] where a is first term of series and r the common difference. \[\Rightarrow \] \[\frac{1}{{{k}_{e}}}=\frac{1}{k}\times \frac{1}{\left( 1-\frac{1}{2} \right)}=\frac{2}{k}\] \[\Rightarrow \] \[{{k}_{e}}=\frac{k}{2}\]
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