## Question

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 ms^{–2.} ^{ }

### Solution

0.005 J

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

The angular frequency of horizontal oscillations is

This amplitude is *A* = 0.01 m. Therefore, total energy is

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