Capacitors problem

4) if angular velocity remains constant ,

t = (2m-1m)/1m/s =1s

θ = ωt = π/2

hence m rotates by an angle 90⁰ .
let θ =0 at t=0

initial vel = (-1)i +(2ω)j

final vel = (-ω)i + (-1)j

change in vel = (1-ω)i + (-1-2ω)j

impulse = m*Δv

Hmm. Thanks. Trying the first one?

2) is it CV3 * cos(wt₀ ) w→√3LC

Initially when battery is connected,charge accumulates and energy is stored.
E_initial=1/2(C/3)V²
Charge on each capacitor=CV/3

Now when, the inductor is connected,energy oscillates.Assume the capacitors behave like a single capacitor,with capacitance C/3.Now when the switch is switched on,the capacitance of the system reduces to C/2.Let at this instant charge on each capacitor be Q.
We can also note that as soon as the 2 plates of the middle capacitor are connected,it immediately discharges,and energy is lost.(ie 1/3rd of original energy).
By energy conservation:
2/3E_initial=E_final
2/3*(1/2(C/3)V²)=1/2*Q²/(C/2)
Hence,
Q=CV/3 [after so much calc. Q is same as initial,proof of the fact that when battery is removed Q remains same]

Therefore amplitude of charge oscillation on the capacitors=CV/3
w=√2/LC [net capacitance of system is C/2]

is the answer given by swaraj correct?
'cos im getting a different one

I don't have the answer, but what is given by Swaraj is in the options!