Classical Mechanic problem No.1

Let the SHM be governed by the equation:

$x(t) = A \sin\left( \omega t + \phi\right) \ ; \ \dot{x} (t)= A \omega \cos\left( \omega t + \phi\right)$

Let $x = 2 \ \mathrm{cm}$ at time $t=t_1$ and $x = 6 \ \mathrm{cm}$ at time $t=t_2$ . That gives:

$\frac{2}{100} = A \sin\left( \omega t_1 + \phi\right) \ ; \ \frac{6}{100} = A \sin\left( \omega t_2 + \phi\right)$

Now, the KE of the particle is:

$\mathcal{T}(t) = \frac{mA^2\omega^2 \cos^2\left( \omega t + \phi\right)}{2}=\frac{m\omega^2 \left(A^2 - A^2\sin^2\left( \omega t + \phi\right) \right)}{2} = \frac{m\omega^2 \left(A^2 - x(t)^2 \right)}{2}$

$\mathcal{T}(t_1)= \frac{m\omega^2 \left(A^2 - x(t_1)^2 \right)}{2} \implies 0.48 = \frac{m\omega^2 \left(A^2 - \left(\frac{2}{100}\right)^2 \right)}{2}$ $\mathcal{T}(t_2)= \frac{m\omega^2 \left(A^2 - x(t_2)^2 \right)}{2} \implies 0.32 = \frac{m\omega^2 \left(A^2 - \left(\frac{6}{100}\right)^2 \right)}{2}$

Dividing both equations and solving for $A$ . Leaving out final simplifications:

$A = 10 \ \mathrm{cm}$