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Let `T_(1)` and `T_(2)` be the time periods of two springs A and B when a mass m is suspnded from them seperately. Now both the springs are connected in parallel and same mass m is suspended with them. Now let T be the time period in this position. Then -A. `T=T_(1)+T_(2)`B. `T=(T_(1)T_(2))/(T_(1)+T_(2))`C. `T^(2)=T_(1)^(2)+T_(2)^(2)`D. `1/T^(2)=1/T_(1)^(2)+1/T_(2)^(2)` |
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Answer» Correct Answer - 4 `T_(1)=2pisqrt(m/k_(1))rArrk_(1)=(4pi^(2)m)/(T_(1)^(2))` `T_(2)=2pisqrt(m/k_(2))rArrk_(2)=(4pi^(2)m)/(T_(2)^(2))` Now `T=2pisqrt(M/k) k=(4pi^(2)m)/(T^(2))` in parallel k=`K_(1)+k_(2)` |
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