# Pendulum spring: obtain expression for maximum speed

We have the following linear 2nd order differential equation that describes its harmonic motion of a pendulum spring:

$$y'' + \frac{k\cdot y}{m} = 0$$

where $m$ is the mass of the pendulum ball, $k$ is the proportionality factor from Hook's law, and $y$ is the elongation from equilibrium position. A possible solution is

$$y(t) = y_{max} \sin{(\frac{2 \pi}{T}t + \rho})$$

How to derive an expression for the maximum speed?

Attempt:

I start by differentiating $y(t)$ to get $v(t)$:

$$v(t) = \frac{2\pi}{T}y_{max} \cos\left( \frac{2\pi}{T}t + \rho \right)$$

Then I optimize $v(t)$ by settings its derivative equal to zero: $\frac{dv(t)}{dt} = 0$

$$\frac{dv(t)}{dt} = - \frac{4 \pi^2}{T^2}y_{max} \sin \left( \frac{2\pi}{T}t + \rho \right) = 0$$

However, I am not sure how to go on from here to actually obtain an expression for the maximum speed. Intuitively I understand that the speed is highest when the pendulum reaches the equilibrium position, when $y=0$, or equivalently when $a=0$ - of course, the acceleration is zero at $y_{max}$ also, so perhaps I should use the two conditions that both $a=y=0$, somehow?

The maximum speed is just the amplitude of $v\left(t\right)$, i.e. $\dfrac{2\,\pi}{T}\,y_{\text{max}}$, and you will get it the first time if the argument of the cosine equals to 0.