A cart of mass 310 g is placed on a frictionless horizontal air track. A spring

Question

A cart of mass 310 g is placed on a frictionless horizontal air track. A spring having a spring constant of 9.0 N/m is attached between the cart and the left end of the track. The cart is displaced 5.0 cm from its equilibrium position.

A cart of mass 310 g is placed on a frictionless horizontal air track. A spring having a spring constant of 9.0 N/m is attached between the cart and the left end of the track. The cart is displaced 5.0 cm from its equilibrium position. Find the period at which it oscillates. Find its maximum speed. Find its speed when it is located 2.0 cm from its equilibrium position.

Explanation / Answer

a) Since the table is frictionless, the only horizontal force on the cart is due to the spring.
F = ma = -kx [1]
m = mass of cart
a = acceleration of cart
k = spring constant
x = position of cart from equilibrium position at a given time

Where's time in this equation, you might ask? It turns out that acceleration and position are functions of time. Specifically, velocity is the first time derivative of position (v(t) = x') and the acceleration is the second time derivative of position (a(t) = x"), so [1] becomes a differential equation with time as the independent variable:
mx" = -kx [2]

I'll cut to the chase here and just tell you that the solution to the differential equation [2] (look up "nonhomogeneous second order linear differential equations with constant coefficients" if you want to know how to solve) with the initial conditions (vo = 0 and xo = 0.045 m) is:
x = xo*cos(wt) [3]
v = -w*xo*sin(wt) [4]
where w = sqrt(k/m) = 6.16 s^-1

a) The frequency of oscillation is w/(2*pi); the period being the reciprocal, 2*pi/w.

2*pi/6.16 = 1.0 s

b) The maximum speed would be the largest value v [4] can have:
max v = w*xo
max v = 6.16*0.045
max v = 0.28 m/s

c) When the cart is 14 cm from the left of the track, it would be 2 cm to the right of the equilibrium position. Using [3] to solve for t:
0.02 = 0.045*cos(wt)
t = 0.18 + 2*pi*n/w (moving to the left) or
t = (2*pi/w - 0.18) + 2*pi*n/w (moving to the right)

Use t = 0.18 s and [4] to solve for v:
v = -6.16*0.045*sin(6.16*0.18)
v = -0.25 m/s

Or use t = 2*pi/w - 0.18 and [4] to solve for v:
v = -6.16*0.045*sin[6.16*(2*pi/6.16 - 0.18)]
v = +0.25 m/s

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