1.

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 )

Answer»

12s
2.4 s
6 s
1.2s

Solution :ABCD
`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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