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The kinetic energy of a particle, executing S.H.M., is 16 J when it is in its mean position. If the amplitude of oscillation is 25 cm, and the mass of the particle is 5.12 kg, the time period of its oscillations is : |
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Answer» `pi//5` s `(1)/(2)m omega^(2)=(16)/(A^(2))=(160000)/(25xx25)` `(1)/(2)xx5.12omega^(2)=256` `omega^(2)=(512)/(5.12)=100` `omega=10" or "(2pi)/(T)=10` `T=(pi)/(5)` s. CORRECT choice is (a). |
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