A linear harmonic oscillator of force constant 2×106 Nm-1and amplitude 0.01 m has a total mechanical energy 160 J. Among the following statements, which are correct?i. Maximum PE is 100 Jii. Maximum KE is 100 Jiii. Maximum PE is 160 Jiv. Minimum PE is zero
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a
Both (i) and (iv)
b
Both (ii) and (iii)
c
Both (i) and (ii)
d
Both (ii) and (iv)
answer is B.
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Detailed Solution
Total mechanical energy is ET = 160 J ∴ Umax = 160 JAt extreme position KE is zero. Work done by spring force from extreme position to mean position isW = 12kA2∴ Kmax = W = 12(2×106)(0.01)2 = 100 J∴ Umin= 160-100 = 60 J