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

Question

A 21.0kg 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= 0.350m, then lets go. The mass undergoes simple harmonic motion with a period of 5.00s. What is the position of the mass 3.950s after the mass is released?

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

Solution: Given, T = 5s, m = 21 kg, A = 0.350 m. For finding position of the oscillating mass at any arbitrary time t, we can use the general relation x = A cos ?t.

Here, ? = 2?/T = 1.256 rad/s. Substituting values, we get x = 0.3487 m for t = 3.95 s.

Magnitude of maximum acceleration can be expressed as fmax = ?2A = 0.552 rad/s2.

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