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 $g=10m{s}^{-2}$

• A. 0.005 J
• B. 0.05 J
• C. 0.5 J
• D. 5 J

Answer 1

The correct option is A 0.005 J

Force acting on spring = mg = 0.5 × 10 = 5N. This force extends the spring by 0.05 m. Therefore, the force constant is
$k=\frac{5}{0.05}=100N{m}^{-1}$
The angular frequency of horizontal oscillations is
$\omega =\sqrt{\frac{k}{m}}=\sqrt{\frac{100}{1.0}}rad{s}^{-1}$
The amplitude is A = 0.01 m. Therefore, total energy is
$E=\frac{1}{2}m{A}^2{\omega }^2=\frac{1}{2}\times 1.0\times {\left(0.01\right)}^2\times {\left(10\right)}^2=0.005J$
Hence the correct choice is (a).