天同師兄,further question

7. You cannot isolate the two beads, as P and Q are connected together by a light rod, the motion of Q is affected by the motion of P.

Hence, you need to consider the rod and two beads as a "single object".
Loss of PE = mgL + mg(2L) = 3mgL
Gain in kintec energy = (1/2)m(vp)^2 + (1/2)m(vq)^2
[where vp and vq are the velocities of P and Q respectively
hence, gain in KE = (m/2).(vp^2 + vq^2)
but vp = Lw and vq = (2L)w, where w is the angular speed
i.e. vp = vq/2
[note: from here, you could see that the loss of PE of Q is 2 times that of P, but the gain in KE of Q is 4 times that of P. That is, the KE of Q comes from (4/5) of the total loss of PE of the "single object", which is 12mgL/5].

equation loss of PE and gain in KE for the "single object"
3mgL = (m/2)((vq/2)^2 + vq^2)
solve for vq gives 24/(5gL)
thus, acceleration = vq^2/(2L) = 12g/5

4. O...sorry, I have missed out that point.You are right that we need to calculate the frequnecy of light using the light speed in the medium, which is 3x10^8/1.5 m/s
hence, frequency = (3x10^8/1.5)/4x10^-7 Hz = 5 x 10^14 Hz
The answer should be option C.