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?

Respuesta :

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.

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