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: (1990)
$ T = 2\pi \sqrt{\frac{m}{k_1 - k_2}} $
$ T = 2\pi \sqrt{\frac{mk_1k_2}{k_1 + k_2}} $
$ T = 2\pi \sqrt{\frac{m}{k_1 + k_2}} $
$ T = 2\pi \sqrt{\frac{m(k_1 + k_2)}{k_1k_2}} $
Solution:
For two springs connected in series, the equivalent spring constant is $ k_{eq} = \frac{k_1k_2}{k_1 + k_2} $. Substituting this into the time period formula $ T = 2\pi \sqrt{\frac{m}{k_{eq}}} $ yields $ T = 2\pi \sqrt{\frac{m(k_1 + k_2)}{k_1k_2}} $.
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