JEE Mains · Physics · STD 11 - 13. oscillations
A spring whose unstretched length is \(\ell \) has a force constant \(k\). The spring is cut into two pieces of unstretched lengths \(\ell_1\) and \(\ell_2\) where, \(\ell_1 = n\ell_2\) and \(n\) is an integer. The ratio \(k_1/k_2\) of the corresponding force constants, \(k_1\) and \(k_2\) will be
- A \(n\)
- B \(\frac{1}{n^2}\)
- C \(n^2\)
- D \(\frac{1}{n}\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{n}\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{l} {k_1} = \frac{C}{{{\ell _1}}}\\ {k_2} = \frac{C}{{{\ell _2}}}\\ \frac{{{k_1}}}{{{k_2}}} = \frac{{C{\ell _2}}}{{{\ell _1}C}} = \frac{{{\ell _2}}}{{n{\ell _2}}} = \frac{1}{n} \end{array}\)
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