InterviewSolution
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InterviewSolution
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A particle performing S.H.M. has a velocity of 10 m//s, when it crosses the mean position. If the amplitude of oscillation is 2 m, then what is the velocity at midway between mean position and extreme position ? |
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Answer» SOLUTION :`v=omegasqrt(a^2-x^2) = omegasqrt(a^2-(a/2)^2)` ` THEREFORE x=a/2 ` `=omegasqrt((3a^2)/4)=(aomegasqrt3)/2=v_max/2sqrt3=(10sqrt3)/2` ` therefore v=5sqrt3 m/s.` |
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