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. Somebod pushes the mass to the position x= 0.350 m, then lets go. The mass undergoes simple harmonic motion with a period of 4.20 s. What is the position of the mass 3.486 s after the mass is released? postton Submit Answer Tries 0/5 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? Submit Answer Tries 0/5

Explanation / Answer

The position of a particle in SHM can be written x = A cos(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=2f. Freq=1/T (T is period) so w= 2/T = 2pi/4.2 = 0.476

So x = A cos(wt) = 0.35 cos (0.476 x 3.486) = 0.1686m .

Acceleration is the second time differential of position, so do x'' = 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. So

Aw^2 = 0.35 x (0.476)2 = 0.7827 m s^-2

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