A 37.0 kg block at rest on a horizontal frictionless air track is connected to t
ID: 1894059 • Letter: A
Question
A 37.0 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 3.30 s.
a) What is the position of the mass 2.6400 s after the mass is released?
b) What is the magnitude of the maximum acceleration the mass undergoes during its motion?
(Consider the same mass and spring discussed in the previous problem)
Explanation / Answer
0.35 m is the maximum amplitude, A, with respect to the reference position.
Using x = A cos t, for t = 0, x(t=0) = A = 0.35 m.
T=3.3 s, and =2/T =2/3.3 =1.9 rad/s
So to find the position at t = 2.640 s.
X =Acost
X=0.35cos(1.9*2.64) =0.108 m
The acceleration is simply the second derivative of the position function i.e X
a = d2x/dt2 = -A2cos(t)
here we are looking for maximum acceleration .so taking cos(t) =1
a = -A2 = -0.35*(1.9)2
a = 1.2635m/s2