In equilibrium
mg =kx
find k
then put value of k in time period formula

Introduction:
When a mass is suspended from a vertical spring, it causes the spring to stretch or elongate. This elongation can be measured as the change in length of the spring. In this scenario, a mass of 1kg is suspended from a vertical spring, causing the spring to increase in length by 0.98m.

Calculating the spring constant:
To find the time period of vibration, we first need to determine the spring constant of the spring. The spring constant (k) is a measure of how stiff the spring is and is given by Hooke's Law:

F = -kx

Where F is the force exerted by the spring, k is the spring constant, and x is the displacement from the equilibrium position. In this case, the mass is pulled downward by gravity, so we can write:

mg = kx

Where m is the mass, g is the acceleration due to gravity, and x is the elongation of the spring. Rearranging the equation, we get:

k = mg/x

Given that m = 1kg and x = 0.98m, we can calculate the spring constant:

k = (1kg)(9.8m/s^2)/(0.98m) = 10 N/m

Calculating the time period of vibration:
The time period of vibration (T) is the time taken for one complete cycle of oscillation. It is given by the formula:

T = 2π√(m/k)

Substituting the values of m and k, we get:

T = 2π√(1kg/10 N/m)

Simplifying the equation, we find:

T ≈ 2π√0.1 s ≈ 2π(0.316) s ≈ 1.99 s

Therefore, the time period of vibration is approximately 1.99 seconds.

Explanation:
When the mass is suspended from the vertical spring, it elongates the spring due to the force of gravity. The elongation is given as 0.98m. By using Hooke's Law and the equation for the spring constant, we can determine the spring constant to be 10 N/m. The time period of vibration is then calculated using the formula for the time period of a simple harmonic oscillator. The resulting time period is approximately 1.99 seconds.