Understanding Time Period of Mass-Spring Vertical Oscillations

When a mass is hung from a spring, it causes the spring to extend. At the equilibrium position, the downward force due to gravity on the mass is balanced by the upward restoring force exerted by the spring.

Let:

• \(m\) be the mass hung from the spring.
• \(g\) be the acceleration due to gravity.
• \(x\) be the extension produced in the spring.
• \(k\) be the spring constant of the spring.

According to Hooke's Law, the restoring force exerted by the spring is proportional to the extension. At equilibrium, the magnitude of the restoring force is \(F_{\text{spring}} = kx\). The gravitational force acting on the mass is \(F_{\text{gravity}} = mg\).

At the equilibrium position, these forces are balanced:

\(F_{\text{spring}} = F_{\text{gravity}}\)

\(kx = mg\)

From this equilibrium condition, we can find the ratio of the mass to the spring constant:

\(\frac{m}{k} = \frac{x}{g}\)

When the mass is set into vertical oscillations, it performs simple harmonic motion (SHM). The time period (\(T\)) of vertical oscillations for a mass-spring system is given by the formula:

\(T = 2\pi \sqrt {\frac{m}{k}}\)

Now, we can substitute the expression for \(\frac{m}{k}\) derived from the equilibrium condition into the formula for the time period:

\(T = 2\pi \sqrt {\frac{x}{g}}\)

This formula shows the time period of oscillation in terms of the extension \(x\) produced by the mass and the acceleration due to gravity \(g\).

Comparing this derived formula with the given options, we find that it matches Option 1.

The final expression for the time period of vertical oscillation is:

\(T = 2\pi \sqrt {\frac{x}{g}}\)