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) Using the equation for the period of an oscillating system, we have:
T = 2π/ω
Where T is the period and ω is the angular frequency. Rearranging the equation, we get:
ω = 2π/T = 2π/1.5 s = 4.1888 rad/s
To find the amplitude A, we can use the equation for the displacement of an oscillating system at any time t:
x(t) = A*cos(ω*t + ϕ0)
At t = 0 s, x(0) = 5.00 cm. Substituting into the equation, we get:
5.00 cm = A*cos(0 + ϕ0) = A*cos(ϕ0)
At t = 0 s, the glider is moving to the right, so the velocity is positive. The velocity is given by the derivative of the displacement equation:
v(t) = -A*ω*sin(ω*t + ϕ0)
At t = 0 s, v(0) = 36.3 cm/s. Substituting into the equation, we get:
36.3 cm/s = -A*ω*sin(0 + ϕ0) = -A*ω*sin(ϕ0)
To find ϕ0, we can divide these two equations:
(5.00 cm)/(36.3 cm/s) = (-A*cos(ϕ0))/(A*ω*sin(ϕ0))
(5.00 cm)/(36.3 cm/s) = -cos(ϕ0)/(ω*sin(ϕ0))
(5.00 cm)/(36.3 cm/s) = (-1/ω)*cot(ϕ0)
cot(ϕ0) = -(5.00 cm*ω)/(36.3 cm/s) = -0.1148
ϕ0 = arccot(-0.1148) = 1.7171 rad
Therefore, ω = 4.1888 rad/s, A = 5.00 cm, and ϕ0 = 1.7171 rad.
(b) The position-time graph for this motion will be a cosine wave with an amplitude of 5.00 cm. The x-axis represents time (in seconds) and the y-axis represents displacement (in centimeters). Each complete cycle will have a duration of 1.5 seconds. The equilibrium position will be at the origin. Two cycles of the oscillation would look like this:
/\
/ \
/ \
/ \
----/--------\-------------
T/2 T/2
(c) At t = 0.5 s, the phase is given by:
ϕ = ω*t + ϕ0 = 4.1888 rad/s * 0.5 s + 1.7171 rad = 3.1155 rad
At t = 1.0 s, the phase is given by:
ϕ = ω*t + ϕ0 = 4.1888 rad/s * 1.0 s + 1.7171 rad = 5.9060 rad