Why is the time period of spring block independent of initial momentum/ energy fuelled into the system? I was recently doing a question on a spring block system and I was just suddenly weirded out by the fact that regardless of how much momentum or energy the block starts with (indirectly how much the spring is compressed), the time period is independent of it. Why does this happen intuitively?
Reason why I think that time period is independent:
$$ T= 2  \pi \sqrt{\frac{ k}{m}}$$
Now ,$ k$ is property of material of spring and $m$ is of the block attached, so clearly it must be independent of initial conditions of the system.
 A: 
I mean, if you compress a spring more like push it more in, then it would have more distance to travel and hence need more time

Intuitively you can think like this : If you compress a spring greater , and then release it the force on the block by spring is greater. Hence it's acceleration(avg)  is faster due to which it can cover more distance in the same time.
A: An analytical explanation with classical mechanics may be useful to some
\begin{array}{l}
T = \frac{1}{2}m{{\dot x}^2}{\rm{   (kinetic)}}\\
V = \frac{1}{2}k{x^2}{\rm{   (potential)}}
\end{array}
Lagrangian approach
\begin{array}{l}
L = T - V\\
\frac{d}{{dt}}\left( {\frac{{\partial L}}{{\partial \dot x}}} \right) - \frac{{\partial L}}{{\partial x}} = 0\\
\ddot x =  - \frac{k}{m}x\\
x = A\cos \left( {\omega t} \right)
\end{array}
This would tell me that the solution is periodic and that the period does not depend on the initial conditions.
Hamiltonian approach
\begin{array}{l}
H\left( {x,{p_x}} \right) = T + V\\
{p_x} = \frac{{\partial L}}{{\partial \dot x}} = m\dot x\\
H = \left( {\frac{1}{{2m}}} \right)p_x^2 + \left( {\frac{1}{2}k} \right){x^2}\\
\dot x = \frac{{\partial H}}{{\partial {p_x}}} = ...\\
{{\dot p}_x} =  - \frac{{\partial H}}{{\partial x}} = ...
\end{array}
From this it would be clear that the motion of the spring in phase space is a circular motion because H is constant. If p is the y-axis and x is the x-axis, you can imagine this circular motion. You'll find from the equations of motion that the tangential velocity on this circle depends linearly on how much momentum you give it. If you give it more momentum, the tangential velocity will be bigger, but the angular velocity will be unchanged.
If you need further convincing, you can easily model this yourself using the time evolution
$$\left( {x,{p_x}} \right) \to \left( {x + \tau \frac{{\partial H}}{{\partial {p_x}}},p - \tau \frac{{\partial H}}{{\partial x}}} \right)$$
This should give you at least three valid intuitive reasons why the initial condition (e.g. momentum) would not change the period.
A: It follows from the spring's force law: $F = kx$, or $a = (k/m)x$. The important point is acceleration is proportional to distance from the center.
You are aware that one solution to the equation is $x = A_1sin(t/T)$
So why is $x = A_2sin(t/T)$ also a solution?
Physicists will answer because both solve $F = kx$. This is correct, but it doesn't give any physical insight. In general, physicists think this way because equations get you farther than physical insight. Particularly when physics gets counter intuitive and highly abstract. But insight is helpful when you can get it.
Here, we can get it. Let us consider the concrete example where $A_2 = 2A_1$. At every instant, solution 2 is twice as far from the center as solution 1. The spring law tells us the force pulling the mass toward the center is twice as big.
You also know the velocity is $x = (A_{1 or 2}/T)cos(t/T)$. The velocity of solution 2 is twice as big as for solution 1 at every instant. This should not be a surprise. For a short time interval $\Delta t$, $\Delta v = a \Delta T$. Starting from $v=0$ at a peak, a short time later, solution 2 has twice the velocity of solution 1. After another short time, solution 2 has gained twice as much more velocity as solution 1.
So in one cycle, the mass in solution 2 travels twice as far at twice the velocity. It should not be surprising that it takes the same time to do so.
A: Roughly speaking, the reason the time period of a spring-block system is independent of initial displacement is that when the displacement is high, even though the block has to travel greater distance to reach the equilibrium position, the force becomes larger with more displacement (and hence the acceleration) in such a way that the block is able to complete the oscillation with the same time period as that of low displacement.
Note that the $\frac{1}{2}kx^2$ potential is very special in this respect. If the potential was $\frac{1}{2}kx^4$, for example, the time period would actually decrease with amplitude because the force at low displacements would be so low (as the slope of potential is almost zero) that it would actually take longer to complete one oscillation even though less distance has to be travelled.
