An air-track glider attached to a spring oscillates with a period of 1.5 s. At t = 0 s the glider
is 5.00 cm left of the equilibrium position and moving to the right at 36.3 cm/s.
(a) Find ω, A, ϕ0.
(b) Draw an appropriately labelled position-time graph for this motion showing 2 cycles.
(c) What is the phase at t = 0.5 s and 1.0 s?
(a) The period of an oscillating system is given by the formula T = 2π/ω, where T is the period and ω is the angular frequency. In this case, T = 1.5 s, so we can solve for ω:
1.5 s = 2π/ω
Cross multiplying, we get:
ω = 2π/1.5 s
= 4.18879 radians/s
The amplitude, A, is the maximum displacement of the glider from the equilibrium position. In this case, the glider is 5.00 cm left of the equilibrium position. Since the motion is symmetric, we can assume that the amplitude is 5.00 cm.
The phase constant, ϕ0, is the initial phase angle. At t = 0 s, the glider is 5.00 cm left of the equilibrium position and moving to the right at 36.3 cm/s. This means that at t = 0, the glider is at its maximum displacement to the left and is in the process of moving to the right. Therefore, the phase angle is 0.
So, ω = 4.18879 radians/s, A = 5.00 cm, and ϕ0 = 0.
(b) Here is an appropriately labeled position-time graph for this motion showing 2 cycles:
^
| ____
| / \
| / \
| __/____________\_______
| T/4 T/2
0 ----- T/4 ------- T/2 ------- 3T/4 ----- T
On the x-axis, we have time (t), and on the y-axis, we have the position (x) of the glider. The graph shows the glider's position as a function of time for 2 cycles.
(c) The phase at t = 0.5 s can be found using the formula:
ϕ = ωt + ϕ0
Substituting the given values, we have:
ϕ = (4.18879 radians/s)(0.5 s) + 0
= 2.094 radians
So the phase at t = 0.5 s is 2.094 radians.
Similarly, the phase at t = 1.0 s is:
ϕ = (4.18879 radians/s)(1.0 s) + 0
= 4.18879 radians
So the phase at t = 1.0 s is 4.18879 radians.