A linear harmonic oscillator of force constant $2 \times 10^{6} N m^{- 1}$ and amplitude 0.01 m has a total mechanical energy 160 J. which of the following statements are correct?
(i) Maximum PE is 100 J (ii) Maximum KE is 100 J
(iii) Maximum PE is 160 J (iv) Minimum PE is zero

Both (i) and (iv)

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Both (ii) and (iii)

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Both (i) and (ii)

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Both (ii) and (iv)

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Solution

The correct option is B Both (ii) and (iii)
Total mechanical energy is $E_{T} = 160 J$
$∴ U_{m a x} = 160 J$
At extreme position KE is zero. Work done by spring force from extreme position to mean position is $W = \frac{1}{2} k A^{2}$
$∴ K_{m a x} = W = \frac{1}{2} (2 \times 10^{6}) (0.01)^{2} = 100 J$
$∴ U_{m i n} = 160 - 100 = 60 J$

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