Question

When a mass of 0.5 kg is suspended from the free end of a spring, it stretches the spring by 0.2 m. This mass is removed and 0.25 kg mass is attached to the same free end of the spring. If the mass is pulled down and released, what is the time period of oscillations?(g= 10 $m{s}^{-2}$)

Answer 1

When 0.3 kg mass is suspended , extension of the spring is 0.2 m.
Since in equilibrium mg = kx$⟹0.5\times 10=k\times 0.2\Rightarrow k=25N/m$

The time period of spring block system is

$T=2\pi \sqrt{\frac{m}{k}}=2\pi \sqrt{\frac{0.25}{25}}=0.63S.$