Comparing two identical pendulums?

[mot]

Homework Statement

Two identical simple pendulums, of length 0.25 m and period T, are mounted side-by-side. One is released from rest with an initial angular displacement of π/9 rad, and the other is started with an initial angular velocity of 0.1 rad/s at θ = 0. They are both started in motion simultaneously. Which of the following statements are true?
A: The equation of motion for both oscillators reads: x = Acos(ωt+δ)
B: After 0.6 seconds they differ in phase by π.
C: After a time of T, they both return to their initial displacement.
D: They both have the same amplitude.

Homework Equations

x(t)=Acos(ωt+δ)

The Attempt at a Solution

What I'm really confused about is the initial angular velocity for the second pendulum. How does this work exactly? Like how do I incorporate it into my equations? The "natural" ω would be √(g/L)≈6.3rad/s. So is the second one driven?
Anyways
A: True this is the equation of motion for all pendulums, right? Or is it False because for the second one, it's "pushed" at first?
B: True. δ=0 for the first one since its initial position is it's amplitude (at t=0) and δ=pi for the second one since sinpi=0, so they always differ in phase by pi (unless that initial angular speed messes that up)
C:True? since this is the definition of a period. Or is it false, again because of the initial angular speed.
D: I don't know how to figure this out, except that the first has one of pi/9rad

 

[mfb]

ω is not the same as the angular velocity in the problem statement. ω is the "velocity" of the phase of your oscillation, the given angular velocity is the change in the position ($\dot \theta$).

B: True. δ=0 for the first one since its initial position is it's amplitude (at t=0) and δ=pi for the second one since sinpi=0, so they always differ in phase by pi (unless that initial angular speed messes that up)

Where does sin(pi) come from now? The second pendulum starts at θ = 0, this is a position.

A and C are true, right.

D: You know the velocity of the second pendulum at a specific point (well, you can calculate it). What is the equation for the velocity?

 

[Simon Bridge]

A: The equation of motion for both oscillators reads: x = Acos(ωt+δ)
... this is true only under a specific condition. Does that condition hold for both pendulums.

B: After 0.6 seconds they differ in phase by π.
... the only way their relative phase changes with time is if they have different frequencies: do they?

C: After a time of T, they both return to their initial displacement.
... that is the definition of the period, and they must return to their initial displacement after some time ... but are their periods actually the same? (JIC)

D: They both have the same amplitude.
... you have to plug the numbers into the equations.

If A is true, then you can write the two pendulums out as:
pendulum 1: $x_1=A_1\cos\theta_1:\theta_1=\omega_1t+\delta_1$

... similarly for pendulum 2.

From B and C you can work out ω.

From the initial conditions you should be able to work out if the amplitudes are the same.