Particle in simple harmonic motion

Question

I am given that a particle in simple harmonic motion has equation of motion $$m\ddot x=-kx$$

I'm told that the particle is at rest at $x=l$ at $t=0.$

I am asked to find the maximum acceleration of the particle. I know that this occurs when the displacement is a maximum but I don't know how to work with this to find the acceleration.

Answer 1

Per the comment by @JohnDouma, it’s clear from inspection of the differential equation that maximum acceleration occurs when the particle’s displacement from the origin is at a maximum. The particle is subject to a symmetric central force, so if it is at rest at $x = l$, then we must have $|x|\le |l|$ for all values of $t$. Thus, maximum acceleration occurs at $|x|=|l|$, and its magnitude is $\frac{k|l|}m$.

Answer 2

equation of motion is $m\ddot x+kx=0$ which is second order linear differential equation which constant coefficients. solve equation and get( equation is homogeneous and it's solution is well known): $$x=A\sin(\sqrt{\frac{k}{m}}t)+B\cos(\sqrt{\frac{k}{m}}t)$$ in which $A$ and $B$ are unknown coefficients. and by using initial value at $x(t=0)=l$ we get $B=0$. also being at rest in the $t=0$ states that $\dot x(0)=0$ so $A=0$. now by double differentiation explicit formula of acceleration as a function of time yields: $$\ddot x(t) = -l\frac{k}{m} \cos(\sqrt{\frac{k}{m}}t)$$
it's maximum occurs when passing through origin and is $l\frac{k}{m}$