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A body is oscillating vertical about its mean position. If its maximum and minimum heights above the surface of the ground are 2 m and 1.5 m, then the speed at the mean position will be `(g=10m//s^(2))`A. `5 m//s`B. `sqrt(2.5)m//s`C. `10 m//s`D. `sqrt(10) m//s` |
Answer» Correct Answer - B `L=h_("max")-h_("min")` `=2-1.5` `2A=0.5" "therefore A=0.25` `v_(m)=Aomega` `=0.25sqrt((g)/(A))` `=sqrt((10xx0.25xx0.25)/(0.25))=sqrt(2.5)m//s` |
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