A 0.9 kg cart is attached between two horizontal springs, of stiffness 3 N/m

Question

A 0.9 kg cart is attached between two horizontal springs, of stiffness 3 N/m, which are attached at each end of a track. the cart has low friction wheels, so friction is negligible. The relaxed length of each spring is 0.1m

the picture is starts at x =0 the cart is at x = 0.5 and the other end is at x = 1.0m

(all of these answers were counted correct on the homework)

Part 1 what is the force by the left spring on the cart? (-1.2, 0, 0) N

Part 2 what is the force by the right spring on the cart? (1.2, 0, 0)N

Now suppose you displace the cart 0.2m to the left of the equilibrium as shown below.

picture x = 0 cart is now at x =0.3 and other end is x= 1.0m

answer fnet =(1.2, 0, 0) N

now what I need help on

If you release the cart from rest(when it is displaced to the left of equilibrium as shown above), what will be its position 0.1 s later?(use approximation v average = p future / m)

r(t=0.1s) = ( _, 0,0)

now at t = 0.1 what is the net force by the springs on the cart?

what will be the positon of the cart 0.1s at t = 0.2 after being released

Answer 1

Releasing the cart leads to simple harmonic motion. As the cart moves, the two springs with force constants of 3 N/m each should produce the same total force as a single spring with a force constant of 6 N/m. The time dependence of simple harmonic motion is described by sine and cosine functions as follows.

d(t) = d₀ cos(ωt) where

d(t) = displacement of the cart from the equilibrium position (middle of track)

d₀ = d(0) = −0.2 m

ω = √(k/m) where k = 6 N/m and m = 0.9 kg, so ω = √(6/0.9) ≈ 2.582

v(t) = −d₀ω sin(ωt) [obtained by taking the derivative of displacement]

plugging in values our equations become

d(t) = –0.2 cos(2.582t) and

v(t) = (0.2)(2.582) sin(2.582t)

from which we can calculate

d(0.1) = –0.2 cos(2.582(0.1)) = –0.193 m (that is 0.193 m left of the middle of the track)

v(0.1) = (0.2)(2.582) sin(2.582(0.1)) = 0.132 m/s (the positive value indicates moving to the right)

If we're calling the equilibrium position x = 0.5, then x(t) = d(t) + 0.5

and x(0.1) = –0.193 + 0.5 = 0.307

and r(0.1) = (0.307,0,0)

At t = 0.1 s the net force on the cart is –k⋅d(0.1) = -6(–0.193) = 1.158 N

At t = 0.2 s we calculate displacement as follows

d(0.2) = –0.2 cos(2.582(0.2)) = –0.174 m (that is 0.174 m left of the middle of the track)

and x(0.2) = –0.174 + 0.5 = 0.326

and r(0.2) = (0.326,0,0)

Answer 2