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A particle at the end of a spring executes simple harmonic motion with a period `t_(1)` while the corresponding period for another spring is `t_(2)` if the oscillation with the two springs in series is T thenA. `T^(2) =t_(1)^(2)+t_(2)^(2)`B. `T= t_(1)+t_(2)`C. `T^(-1)=t_(1)^(-1)+t_(2)^(-1)`D. `T^(-2)=t_(1)^(-2)+t_(2)^(-2)` |
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Answer» Correct Answer - A `t_(1)=2pi sqrt(m/k_(1))` `Rightarrow k_(1)=(4pi^(2)m)/t_(1)^(2) and k_(2)=(4pi^(2)m)/t_(2)^(2)` `1/k = 1/k_(1) + 1/k_(2)=(t_(1)^(2)+t_(2)^(2))/(4pi^(2)m)` `T = 2pi sqrt(m1/k) = 2pi sqrt(m( t_(1)^(2) + t_(2)^(2))/(4pi^(2)m))` `T^(2)= t_(1)^(2)+ t_(2)^(2)` |
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