Two springs, of force constant k₁ and k₂ are connected to a mass m as shown. The frequency of oscillation of the mass is f. If both k₁ and k₂ are made four times their original values, the frequency of oscillation becomes (a) 2f (b) f/2 (c) f/4 (d) 4f

Answer: (a). Springs on both sides act in parallel: f = 1/2π√((k₁+k₂)/(m)). Both constants × 4: keff becomes 4(k₁+k₂), so f' = √(4) f. So the answer is (a) 2f.

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Spring Systems

Two springs, of force constant $k_1$ and $k_2$ are connected to a mass $m$ as shown. The frequency of oscillation of the mass is $f.$ If both $k_1$ and $k_2$ are made four times their original values, the frequency of oscillation becomes

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Answer: (a)

Springs on both sides act in parallel: $f = \frac{1}{2\pi}\sqrt{\frac{k_1+k_2}{m}}$ .
Both constants $\times 4$ : $k_{eff}$ becomes $4(k_1+k_2)$ , so $f' = \sqrt{4}\,f$ .
So the answer is (a) $2f$ .

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