Question

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?

Answer 1

Answer:

Explanation:

We use the harmonic motion position equation:

$x(t) = A\cos(\omega t+\phi)$

where A = 0.350 and for t = 0

$x(0) A = A\ cos(\phi)$

so: $\phi = 0$

and also:

$\omega = \frac{2\pi}{T} = \frac{2\pi}{4.10} = 1.532 rad/s$

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