A 37.0kg block at rest on a horizontal frictionless air track is connected to th

Question

1. A 37.0kg 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 4.90s. What is the position of the mass 4.018s after the mass is released?
2. 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

m = Given
x0 = Given (displacement from equilibrium)
T = Given
t = Given
x = position at time t = ?
This problem involves using the formula for simple harmonic motion. The mass
is not used at all for the solution. To nd the position at the given time, use the
formula
x = A cos (?t + ?)
A is the amplitude, that is, the displacement from equilibrium or what is called
x0 here.
? = 2*pi/T
and ? is zero in this case because the mass is initially at the greatest displacement.
The reason being is that the mass is brought to an initial displacement and let
go. So the equation becomes,
x = x0 cos( 2*pi*t/T)

x = 0.35cos( 2*pi*t/4.9)

position of mass at t = 4.018 is

x = 0.35cos( 2*pi*4.018/4.9)

= 0.35cos(5.15221)

= 0.42578 m

2. The acceleration is simply the second derivative of the position function. Since
the position function was calculated in the previous problem, simply dierentiate

x" = a = -4pi2/T2x0cos( 2*pi*t/T)

therefore maximum acceleration is mod a = 4pi2/T2x0cos(0)

= 0.57548 m/s2

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