A spring stretches by 0.05 m when a mass of 0.5 kg is hung from it. A body

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A spring stretches by 0.05 m when a mass of 0.5 kg is hung from it. A body of mass 1.0 kg is attached to one of its ends, the other end being fixed to the wall. The body is pulled 0.01 m along a horizontal frictionless surface and released. What is the total energy of the oscillator. Assume the string to have negligible mass and take

g = 10 m s^{- 2}

Options:

A . 0.005 J

B . 0.05 J

C . 0.5 J

D . 5 J

Answer: Option A
:
A
Force acting on spring = mg = 0.5 × 10 = 5N. This force extends the spring by 0.05 m. Therefore, the force constant is

k = \frac{5}{0.05} = 100 N m^{- 1}

The angular frequency of horizontal oscillations is

ω = \sqrt{\frac{k}{m}} = \sqrt{\frac{100}{1.0}} r a d s^{- 1}

The amplitude is A = 0.01 m. Therefore, total energy is

E = \frac{1}{2} m A^{2} ω^{2} = \frac{1}{2} \times 1.0 \times {(0.01)}^{2} \times {(10)}^{2} = 0.005 J

Hence the correct choice is (a).

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