InterviewSolution
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The velocity `v` of a particle of mass in moving along a straigh line change withh time `t` as `(d^(2)v)/(dt^(2)) = - Kv` where `K` is a particle constant which of the folloewing statement is correct?A. The particle does not perform `SHM`B. The particle perform `SHM`with time period `2pi sqrt((m)/(k))`C. The particle perform `SHM`with time frequency`sqrt(k)/(2pi)`D. The particle perform `SHM`with time period `(2pi)/(K)` |
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Answer» Correct Answer - C `(d^(2)v)/(dt) = - kv^(2)` this equation has standard solution `v = v_(0) sin(sqrt(k)t + theta)`where `= omega = sqrt(k)` Hence the particle executes `SHM` with angular frequency `omega = sqrt(k) ` or frequency `f = sqrt(k)/(2pi)`. |
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