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A particle executes `SHM` represented by the equation, `y = 0.02 sin (3.14t+(pi)/(2))m`. Find (i) amplitude (ii) time period (iii) frequecy (iv) epoch (v) maximum velocity and (vi) maximum acceleration. |
Answer» Compare the equation `y = 0.02 sin (3.14t+(•)/(2))` with the general form of the equation, `y = A sin (•t+•)` (i) Amplitude `A = 0.01m` (ii) Timw period is given by `T = (2•)/(•)`or `T=(2•)/(3.14) = 2s` (iii) Frequecny `f = (1)/(T) = (1)/(2) Hz = 0.5 Hz` (iv) Epoch `• =(•)/(2) = (3.14)/(2) = 1.57 rad` Epoch is initial phase (v) Maximum velocity `v_(max) = A• = 0.02 xx 3.14 = 0.0628 ms^(-1)` (vi) Maximum acceleration `a_(max) = A•^(2) = 0.02 xx (3.14)^(2) = 0.197 ms^(-2)` |
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