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A 23.0 kg block at rest on a horizontal frictionless air track is connected to the wall via a spring. The equilibrium position of the mass is defined to be at x=0. Somebody pushes the mass to the position x= 0.350 m, then lets go. The mass undergoes simple harmonic motion with a period of 4.10 s. What is the position of the mass 3.403 s after the mass is released?

Respuesta :

Answer:

Explanation:

We use the harmonic motion position equation:

[tex]x(t) = A\cos(\omega t+\phi)[/tex]

where A = 0.350 and for t = 0

[tex]x(0) A = A\ cos(\phi)[/tex]

so: [tex]\phi = 0[/tex]

and also:

[tex]\omega = \frac{2\pi}{T} = \frac{2\pi}{4.10} = 1.532 rad/s[/tex]

so we have:

x(t)=0.350cos(1.532 t)

For t = 3.403 s

x(3.403)=0.350cos(1.532 (3.403)) = 0.348 m

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