Space Force

I can try to explain this is simplest terms without a single formula.

As the mass increases, with it increases the inertia (of motion) of the spring-mass system, thus making it harder to accelerate ( or decelerate) , thus increasing the time-period of each cycle, causing the frequency to decrease.

Consider a mass-spring system, similar to the one shown in the graphic. Write Newton's Second Law:

$\large{F = m \ddot{x} \\ -k x = m \ddot{x} \\ \ddot{x} = - \frac{k}{m} x}$

Taking two derivatives gives the same function back, but scaled and negated. This means that the solution consists of sine waves (simple harmonic motion).

$\large{x = A \, cos(\omega t) + B \, sin(\omega t) \\ \\ \dot{x} = -A \, \omega \, sin(\omega t) + B \, \omega \, cos(\omega t) \\ \ddot{x} = -A \, \omega^2 \, cos(\omega t) - B \, \omega^2 \, sin(\omega t) = -\omega^2 \, x \\ \omega = \sqrt{\frac{k}{m}}}$

With no astronaut on the spring, the effective mass is the spring's own mass. With the astronaut attached, the effective mass is larger, and the angular frequency is correspondingly smaller (since the mass is in the denominator).

For a bit more fun, let's suppose that at time $t = 0$ , the initial displacement (position) is $D$ , and that the initial speed is zero. This gives us the following equations to solve for position and velocity:

$\large{x_0 = D = A \times 1 + B \times 0 \\ \dot{x}_0 = 0 = -A \, \omega \times 0 + B \, \omega \times 1 \\ A = D \\ B = 0}$

This gives the following expression for x:

$\large{x = D \, cos(\omega t) \\ \omega = \sqrt{\frac{k}{m}}}$

Note that the vibration frequency depends on the mass attached to the spring, but the maximum displacement from the equilibrium position depends only on the initial displacement, regardless of the mass.

The frequency of oscillation of a mass-spring system is inversely proportional to the active compressional mass (in this case). Thus the frequency decreases with increased mass.

Can someone explain how are we commenting on the displacement of the spring??

In my words...when I grasp any moving object it stops moving

The only existing force is the tension of the spring. Using $Newton's\quad second\quad law$ we have $\overset { \rightarrow }{ a } =\frac { \overset { \rightarrow }{ T } }{ m } \Longrightarrow a=\frac { k*x }{ m }$ Since $\overset { \rightarrow }{ T }$ is constant and $k$ also then $x$ the displacement is constant. And as the mass of the system has increased then it's accelaration will decrease and the vilocity also, then the frequence will decrease.