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A particle of mass 10 g is performing SHM. Its kinetic energies are 4.7 J and 4.6 J when the displacements are 4 cm and 6 cm , respectively. Compute the period of oscillation. |
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Answer» SOLUTION :Data : m = 0.01 kg, `KE_(1) = 4.7 J, x_(1) = 4 xx 10^(-2) m, KE_(2) = 4.6 J ` ` x_(2) =6 xx 10^(-2) m` Since the total energy of a PARTICLE is SHM is CONSTANT . ` KE_(1) + PE_(1) = KE_(2) +PE_(2)` ` KE_(1) -KE_(2) = PE_(2) -PE_(1) = 1/2m omega^(2) (x_(2)^(2) -x_(1)^(2))` ` 1/2 m((4pi^(2))/(T^(2)) (x_(2)^(2) -x_(1)^(2)) "" (therefore omega = (2pi)/T)` ` T^(2) = 2mpi^(2) = ((x_(2)^(2)-x_(1)^(2)))/(KE_(1) -KE_(2))` ` = 2(0.01) pi^(2) ([(6^(2)-4^(2)) xx 10^(-4)])/(4.7 - 4.6)` ` = 0.2 pi^(2) xx 20 xx 10^(-4)= 4pi^(2) xx 10^(-4)` The period of oscillation of the particle. ` T= sqrt(4pi^(2) xx 10^(-4)) = 2pi xx10^(-2) = 6.284 xx 10^(-2)S` |
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