For a 2.7kg mass oscillating in simple harmonic motion (spring constant: 360N/m) with an amplitude of oscillations measured at 3.4cm.
How do I calculate the total mechanical energy, maximum speed and at what positive displacement, the mass has half the maximum speed.
For one-dimensional simple harmonic motion, the equation of motion is $$ \ddot x(t)+\omega^2 x(t)=0 $$ where $ \omega = \sqrt{\frac{k}{m}}$, $m$ is the inertial mass of the oscillating body, $x(t)$ is its displacement from the equilibrium position, and $k$ is the spring constant for a mass on a spring.
Solving the differential equation above produces a solution that is a sinusoidal function: $$ x(t) = A\cos\left(\omega t + \varphi\right), $$ where $A$ is the amplitude (maximum displacement from the equilibrium position), $ω = 2πf$ is the angular frequency, and $\varphi$ is the phase.
The velocity and acceleration as a function of time are \begin{align} v(t) &= \dot x(t) = - A\omega \sin(\omega t+\varphi)\\ a(t) &= \ddot x(t)=-\omega^2 x(t)= - A \omega^2 \cos( \omega t+\varphi). \end{align} The maximum speed is $v_{\max}= \omega A$ at equilibrium point and the maximum acceleration $a_{\max}= \omega^2 A$ at extreme points.
The velocity as function of displacement is obtained from \begin{align} v^2(t)&=\left[- A\omega \sin(\omega t+\varphi)\right]^2\\& =\omega^2 A^2\sin^2(\omega t+\varphi)=\omega^2 A^2\left[1-\cos^2(\omega t+\varphi)\right]\\ &=\omega^2 (A^2-x^2(t)) \end{align} that is $$ v(x)=\omega\sqrt{A^2-x^2} $$ So the positive displacement $\tilde x$ at which the mass has half the maximum speed will be found solving $$ v\left(\tilde x\right)=\omega\sqrt{A^2-\tilde x^2}=\frac{1}{2}v_{\max}=\frac{1}{2} \omega A $$ that is $$\omega^2 (A^2-\tilde x^2)= \frac{\omega^2 A^2}{4}\quad \Longrightarrow \tilde x=\frac{\sqrt 3}{2}A$$
The kinetic energy $K$ of the system at time $t$ is $$ K(t) = \frac{1}{2} mv^2(t) = \frac{1}{2}m\omega^2A^2\sin^2(\omega t + \varphi) = \frac{1}{2}kA^2 \sin^2(\omega t + \varphi), $$ and the potential energy is $$ U(t) = \frac{1}{2} k x^2(t) = \frac{1}{2} k A^2 \cos^2(\omega t + \varphi). $$ The total mechanical energy has a constant value $$ E = K + U = \frac{1}{2} k A^2. $$