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?

$prob25a_SpringShort.gif$


$prob25a_SpringLong.gif$

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?

Answer 1

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 f_max = ?²A = 0.552 rad/s².