Question

The position of an object that is oscillating on an ideal spring is given by the equation x = (0.1 m) cos[(1.26s-1)t]. At time t = 0.8 s, what is the magnitude of the acceleration of the object?

Answer 1

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,

$a=\frac {\partial^2x}{\partial t^2}$

$a=\frac {\partial^2(0.1\times cos(1.26t)}{\partial t^2}$

$a=0.1\times 1.26\times 1.26 (-cos(1.26t))$

When t = 0.8 sec,

$a=-0.15876\times (cos(1.26\times 0.8))$

$a=-0.15876\times 0.5336\ ms^{-2}$

$a=-0.0847\ ms^{-2}$

The magnitude of the acceleration is 0.0847 m/s².