InterviewSolution
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InterviewSolution
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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=t_(1)+t_(2)`B. `T^(2)=t_(1)^(2)+t_(2)^(2)`C. `T^(1)=t_(1)^(-1)+t_(2)^(-1)`D. `T^(2)=t_(1)^(-2)+t_(2)^(-2)` |
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Answer» `t_(1)=2pisqrt((m)/(k_(1)))" " ....(1),` `t_(2)=2pisqrt((m)/(k_(2)))" " ....(2),` when springs are in series then `T=2pisqrt(((m)/(k_(1)k_(2)))/(k_(1)+k_(2)))=2pisqrt(m(k_(1)+k_(2)))/(k_(1)k_(2))` squaring and adding (1) and (2) we get `t_(1)^(2)+t_(1)^(2)=4pi^(2)(m)/k_(1)+4pi^(2)(m)/k_(2)` `=4pi^(2)m(k_(1)+k_(2)/(k_(1)k_(2)))` or `t_(1)^(2)+t_(2)^(2)=T^(2)` |
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