A mass m is suspended from the two coupled springs connected in series. The force constant for springs are k₁ and k₂. The time period of the suspended mass will be (a) T = 2π√((m)/(k₁-k₂)) (b) T = 2π√((mk₁k₂)/(k₁+k₂)) (c) T = 2π√((m)/(k₁+k₂)) (d)

Answer: (d). Series combination: keff = (k₁k₂)/(k₁+k₂). T = 2π√mkeff = 2π√((m(k₁+k₂))/(k₁k₂)) So the answer is (d).

A mass m is suspended from the two coupled springs connected in… | NEET 1990 (AIPMT)

Q101NEET 1990 · AIPMTEasy

Spring Systems

A mass $m$ is suspended from the two coupled springs connected in series. The force constant for springs are $k_1$ and $k_2$ . The time period of the suspended mass will be

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Answer: (d)

Series combination: $k_{eff} = \frac{k_1k_2}{k_1+k_2}$ .
$T = 2\pi\sqrt{\frac{m}{k_{eff}}} = 2\pi\sqrt{\frac{m(k_1+k_2)}{k_1k_2}}$
So the answer is (d).

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