(a) 0.0021 s, 2926.5 rad/s
The frequency of the B note is

The time taken to make one complete cycle is equal to the period of the wave, which is the reciprocal of the frequency:

The angular frequency instead is given by

And substituting
f = 466 Hz
We find

(b) 20 Hz, 125.6 rad/s
In this case, the period of the sound wave is
T = 50.0 ms = 0.050 s
So the frequency is equal to the reciprocal of the period:

While the angular frequency is given by:

(c) 
The minimum angular frequency of the light wave is

so the corresponding frequency is

and the period is the reciprocal of the frequency:

The maximum angular frequency of the light wave is

so the corresponding frequency is

and the period is the reciprocal of the frequency:

(d) 
In this case, the frequency is

So the period in this case is

While the angular frequency is given by
