A vertical spring oscillates with period 6 second with mass m is suspended from it. When the mass is at rest, the spring is stretched through a distance of (Take, acceleration due to gravity, g = ^2=10 ~m / s ^2 )

Step-by-step solution

The equilibrium position occurs when the gravitational force mg balances the spring's restoring force kx , giving mg = kx and thus x = mg k . The period of oscillation is T = 2 m k , which can be rearranged as m k = T^2 4 ^2 . Substituting into the expression for x yields x = g T^2 4 ^2 . Given T = 6 s and g = ^2 = 10 m/s², we compute: x = ^2 36 4 ^2 = 36 4 = 9 m. The spring stretches through a distance of 9 m at equilibrium.