Answer:
[tex]\frac{d^{2}x}{dt^{2}}=1.65(m/s^{2})[/tex].
Explanation:
To solve this problem we are going to consider the particle to be moving along the x axis and we are also going use the differential equation of the simple harmonic oscillator:
[tex]\frac{d^{2}x}{dt^{2}}+\omega^{2}x=0[/tex],
[tex]\frac{d^{2}x}{dt^{2}}=-\omega^{2}x[/tex],
where [tex]x[/tex] is the displacement along the x axis from the equilibrium point, and [tex]\omega[/tex] is the angular frecuency.
We know that [tex]\frac{d^{2}x}{dt^{2}}[/tex] is the acceleration of the particle and that the angular frecuancy depends on the period:
[tex]\omega=\frac{2\pi}{T}[/tex].
By substitution we get
[tex]\frac{d^{2}x}{dt^{2}}=-\frac{2\pi}{T}x[/tex].
We get the maximum and minimum acceleration at a displacement equals the [tex]\pm amplitude[/tex]. Since we are moving along the x-axis, we get the maximum acceleration at [tex]x=-0.76m[/tex].
So
[tex]\frac{d^{2}x}{dt^{2}}=-\frac{2\pi}{2.9}*(-0.76)[/tex],
[tex]\frac{d^{2}x}{dt^{2}}=1.65(m/s^{2})[/tex].
