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?
To solve this problem, we will use the equation for simple harmonic motion:
x(t) = A cos(ωt + ϕ0)
where x(t) is the position of the glider at time t, A is the amplitude, ω is the angular frequency, and ϕ0 is the phase constant.
(a) To find ω, we can use the formula:
T = 2π/ω
where T is the period. Plugging in the given period of 1.5 s, we have:
1.5 = 2π/ω
Cross multiplying and rearranging, we get:
ω = 2π/1.5
ω = 4.18879 rad/s
To find A, we use the fact that the glider is 5.00 cm to the left of the equilibrium position. Since cosine is maximum at ωt = ϕ0, we know that at t = 0, cos(ωt + ϕ0) = 1. Thus, we can write:
x(0) = A cos(ϕ0) = -5.00 cm
Since cosine is an even function, we know that cos(ϕ0) = cos(-ϕ0). Therefore, we can write:
A cos(ϕ0) = A cos(-ϕ0) = -5.00 cm
Solving for A, we have:
A = 5.00 cm
To find ϕ0, we can use the given initial velocity. The glider is moving to the right at 36.3 cm/s. At t = 0, we know that the velocity is maximum, so:
v(0) = -Aω sin(ϕ0) = 36.3 cm/s
Since sine is maximum at ωt = π/2 + ϕ0, we can write:
sin(π/2 + ϕ0) = 1
Substituting this into the equation above, we have:
-Aω = 36.3 cm/s
-5.00 cm * 4.18879 rad/s = 36.3 cm/s
ϕ0 = -2.97799 rad
(b) To draw the position-time graph for 2 cycles, we need to find the range of t that corresponds to 2 periods. Since the period is 1.5 s, 2 periods would correspond to 3 s. To draw the graph, we can choose any values of t within this range and plug them into the equation x(t) = 5.00 cm * cos(4.18879t - 2.97799 rad). Here is an example position-time graph:
```
y-axis (position in cm)
| .
| .
| .
| .
| .
| .
| .
| .
| .
| . .
| . .
| . .
| . .
| . .
| . .
| . .
| . .
|aaaaaaaaaaaaaaa|bbbbbbbbbbbbbbb|ccccccccccccccc
0 1 2 3 → x-axis (time in s)
```
In this graph, the glider starts at the far left (positions labeled "a"), moves to the equilibrium position (positions labeled "b"), and then moves to the far right (positions labeled "c"). The repeating pattern represents two cycles of the glider's motion.
(c) To find the phase at t = 0.5 s, we can plug in t = 0.5 s into the equation x(t) = 5.00 cm * cos(4.18879t - 2.97799 rad):
x(0.5) = 5.00 cm * cos(4.18879 * 0.5 - 2.97799 rad)
= 5.00 cm * cos(0.9433 rad)
= 5.00 cm * 0.5985
= 2.9925 cm
Therefore, at t = 0.5 s, the glider is approximately 2.9925 cm to the right of the equilibrium position.
To find the phase at t = 1.0 s, we can plug in t = 1.0 s into the equation:
x(1.0) = 5.00 cm * cos(4.18879 * 1.0 - 2.97799 rad)
= 5.00 cm * cos(1.21149 rad)
= 5.00 cm * 0.3005
= 1.5025 cm
Therefore, at t = 1.0 s, the glider is approximately 1.5025 cm to the right of the equilibrium position.