Question

Two mass m₁ = 2.0 kg and m₂ = 10 kg are suspended by a spring of force constant k = 25Nm^-1. When both are in equilibrium then m₁ is removed, then find out the angular frequency and amplitude of m₂. (g = 10ms_-2)

Answer 1

m₁ = 2.0 kg; m₂ = 1.0 kg; k = 25 N. m^-1

g = 10 m s^-2

After removing m₁ angular frequency of m₂

_{\(\omega = \sqrt{\frac{k}{m_2}} = \sqrt{\frac{25}{1}} = 5\)}

or ω = 5 rad s^-1

After removing m₁, decrease in displacement of the spring amplitude of oscillations m₂.

F = m₁g = ka

\(\therefore\ a = \frac{m_1g}{k} = \frac{2\times10}{25} = \frac{4}{5} = 0.8\ m\)

or a = 0.8 m