InterviewSolution
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InterviewSolution
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Maximum kinetic energy of a particle of mass 1 kg in SHM is 8 J. Time period of SHM is 4 s. Maximum potential energy during the motion is 10 J. ThenA. amplitude of oscillations is approximately 3.53 mB. minimum potential energy of the particle is 4 JC. maximum acceleration of the particle is approximately 6.3 `ms^(-2)`D. minimum kinetic energy of the particle is 2 J |
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Answer» Correct Answer - C Maximum kinetic energy = energy of oscillation in SHM `therefore" "8=(1)/(2)kA^(2)` `therefore" "kA^(2)=16" "...(i)` Further, `2pi sqrt((m)/(k))=4 rArr K = (pi^(2))/(4)" "...(ii)` From Eqs. (i) and (ii), we get `k = 2.4 Nm^(-1)and A = 2.53m` Maximum acceleration of the particle will be `a_(max)=omega^(2)A=(k)/(m)A=((25)/(1))(2.53)=6.3 ms^(-2)` |
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