A 21.0 kg block at rest on a horizontal frictionless air track is connected to t

Question

A 21.0 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 5.30 s.

A. What is the position of the mass 4.452 s after the mass is released?
B. Consider the same mass and spring discussed in the previous problem. What is the magnitude of the maximum acceleration the mass undergoes during its motion?

Explanation / Answer

You should know that the position of a particle in SHM can be written x = Acos(wt+d). d is a phase shift, and since we take t=0 as the point where the spring is stretched most, and we're using a cosine, d=0. A is the amplitude: 0.35m. We need to find w. w=2pf. Freq=1/T (T is period) w= 2p/T = 0.377p So x = A cos(wt) = 0.35 cos (0.377p x 4.452) = 0.346m . a= -A (w^2) cos (wt). The maximum value cos(wt) can take is 1, so set it equal to 1 to find the maximum acceleration. A x w^2 = 0.35 x (0.377p^2) = 1.30m s^-2

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