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A particle performing S.H.M . Undergoes displacement A/2 (where A= amplitude of S.H.M. ) in one second. At t=0 the particle was located at either extreme position or mean position. The time period of S.H.M. can be :( consider all possible cases ) |
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Answer» 12s `oo=(THETA)/(t)` `oo_(1)^(2)=PI//6,(2PI)/(T_(1))=(pi)/(6),T_(1)=12` `oo_(2)=5pi//6,(2pi)/(T_(2))=(pi)/(6),T_(2)=2.4` `T_(1)=12sec` `T_(2)=2.4 SEC` `theta_(1)^(1)=(pi//3)/(1),(2pi)/(T_(1)^(1))=(pi)/(3), T_(1)^(1)=6 sec` `theta_(2)^(1)=(5pi//3)/(1),(2pi)/(T_(2)^(1))=(5pi)/(3), T_(2)^(1)=1.2 sec`
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