Up and down and up and down and...

Using a simple one-dimensional kinematic equation for an object with constant acceleration, we model the ball as a projectile with initial height A and initial velocity 0. Its position in the y-direction as a function of time is now

$y(t) = A + \frac{1}{2}gt^2$

Where g is the gravitational acceleration (-9.8 m/s^2). To find the period T, we need to find when the height is zero, so

$0 = A + \frac{1}{2}gt^2$

$T^2 = \frac{-2A}{g}$

From this equation, it's clear that $T^2$ is proportional to A, so $T$ is proportional to $A^{\frac{1}{2}}$ .

Since the descent of the ball is essentially the reverse of the ascent. This means that we only need to consider the descent to find $w$ . We need to use suvat to find $w$ :

Using:

$s = u \cdot t + \frac{1}{2} \cdot a \cdot t^2$

Let:

$s = A, t = \frac{1}{2}T$ and $u = 0ms^{-1}$

This Gives:

$A = \frac{1}{2} \cdot a \cdot (\frac{T}{2})^2$

$A = \frac{1}{2} \cdot a \cdot \frac{1}{4} \cdot T^2$

$A = \frac{1}{8} \cdot a \cdot T^2$

Since $\frac{1}{8} \cdot a$ is a constant:

$A \propto T^2$

$T \propto A^{0.5}$

Therefore, $\boxed{w = 0.5}$

Since mechanical energy is conserved, at the top of the "bounce" the potential energy is equal to the kinetic energy at the bottom of the bounce. The amplitude A is equal to the relative height, so $mgA = \frac{1}{2}mv^2 \rightarrow v = \sqrt{2gA}$ . Thus, at the bottom the oscillator goes upward at such a speed. The amount of time it takes for it to return is reduced to a simple kinematics problem, and using the fact that $v_f = v_0 - gt$ and that $v_f = -v_0$ , we come to $t = \frac{2v_0}{g} = \frac {2\sqrt{2gA}}{g}$ . Thus, $T$ is proportional to $A^{\frac{1}{2}}$ .

$\frac { A }{ 2 } ={ v }_{ i }+\frac { 1 }{ 2 } g{ (\frac { T }{ 2 } ) }^{ 2 }\Rightarrow A=g\frac { { T }^{ 2 } }{ 4 } \Rightarrow { A }^{ \frac { 1 }{ 2 } }=\frac { \sqrt { g } }{ 2 } T$

$w=0.5$