Answer:
The magnitude of the acceleration is 0.0847 m/s²
Explanation:
Given:
Displacement equation for the spring:
x = 0.1*cos(1.26*t)
Acceleration is the second derivative of displacement w.r.t.
Thus,
[tex]a=\frac {\partial^2x}{\partial t^2}[/tex]
[tex]a=\frac {\partial^2(0.1\times cos(1.26t)}{\partial t^2}[/tex]
[tex]a=0.1\times 1.26\times 1.26 (-cos(1.26t))[/tex]
When t = 0.8 sec,
[tex]a=-0.15876\times (cos(1.26\times 0.8))[/tex]
[tex]a=-0.15876\times 0.5336\ ms^{-2}[/tex]
[tex]a=-0.0847\ ms^{-2}[/tex]
The magnitude of the acceleration is 0.0847 m/s².