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A particle executes SHM with a time period of 2 s and amplitude 5 cm Find (i) displacement (ii) velocity and (ii) acceleration, after 1/3 second ; starting from the mean position. |
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Answer» (i) displacement is x = 5 sin (2π/T t ) = 5 sin (π t ) [cm] x(t=1/3) = 5 sin (π/3) = 5*√3/2. (ii) then the velosity would be v = 5π cos (π t) [cm/s] v(t=1/3) =2.5π (iii) acceleration a = -5 π2sin (π t) a(t=1/3)= -2.5π2√3 |
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